Exercise Set 3F
These questions are considered intermediate to advanced Higher Level.
Question 1
An object is projected from \(o\) with a speed \(u=35\mbox{ m/s}\) at an angle \(A\) relative to the horizontal ground.
If the projectile is to hit a target that is located at \(8\vec{i}+5\vec{j}\mbox{ m}\) from \(o\), what are the two possible values of \(A\)?
\(88.13^{\circ}\) and \(33.82^{\circ}\)
First, an \(x\) component equation.
\begin{align} u_x = 35\cos A && a_x=0 && s_x=8 && t=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt\]
\begin{align}\downarrow\end{align}
\begin{align}8=35\cos A t\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t = \frac{8}{35\cos A}\end{align}
\[\,\]
Now, a \(y\) component equation.
\begin{align} u_y = 35\sin A && a_y=-9.8 && s_y=5 && t=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}5=35\sin A t -\frac{1}{2}(9.8)t^2\end{align}
If we now sub our first equation for \(t\) into this equation, we get
\begin{align}5&=35\sin A\left(\frac{8}{35\cos A}\right) -\frac{1}{2}(9.8)\left(\frac{8}{35 \cos A}\right)^2\\&=8 \tan A-\frac{0.256}{\cos^2 A} \end{align}
Inserting the following trigonometric identity
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
we then obtain
\begin{align}5 = 8 \tan A- 0.256(1+\tan^2 A)\end{align}
Setting \(x=\tan A\), this becomes
\begin{align}5=8x-0.256(1+x^2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.256 x^2-8x+5.256=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{8 \pm \sqrt{(-8)^2-4(0.256)(5.256)}}{2(0.256)}\\ &\approx 3.57 \pm 3.05\end{align}
and therefore \(x=30.58\) or \(x=0.67\).
\begin{align}A&= \tan^{-1}(30.58)\\&\approx 88.13^{\circ}\end{align}
and
\begin{align}A&= \tan^{-1}(0.67)\\&\approx 33.82^{\circ}\end{align}
To answer this, we should find the time taken for the object to reach \(s_x = 8\) and \(s_y=5\) separately before equating those times and solving for \(A\).
To start, we first wish to find \(t\) when \(s_x=8\). We should therefore use an \(x\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 35\cos A && a_x=0 && s_x=8 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}8=35\cos A t\end{align}
Rearranging, this gives
\begin{align}t = \frac{8}{35\cos A}\end{align}
\[\,\]
Next, we should find \(t\) when \(s_y=5\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 35\sin A && a_y=-9.8 && s_y=5 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}5=35\sin A t -\frac{1}{2}(9.8)t^2\end{align}
If we now sub our first equation for \(t\) into this equation, we get
\begin{align}5&=35\sin A\left(\frac{8}{35\cos A}\right) -\frac{1}{2}(9.8)\left(\frac{8}{35 \cos A}\right)^2\\&=8 \tan A-\frac{0.256}{\cos^2 A} \end{align}
Inserting the following trigonometric identity
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
we then obtain
\begin{align}5 = 8 \tan A- 0.256(1+\tan^2 A)\end{align}
Now, let us temporarily set \(x=\tan A\), reducing this equation to
\begin{align}5=8x-0.256(1+x^2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.256 x^2-8x+5.256=0\end{align}
This can be solved using the quadratic formula to obtain
\begin{align}x&=\frac{8 \pm \sqrt{(-8)^2-4(0.256)(5.526)}}{2(0.256)}\\ &\approx 30.58 \pm 0.67\end{align}
and therefore \(x=30.58\) or \(x=0.67\). Inverting this, we therefore obtain for the first value of \(x\)
\begin{align}A&= \tan^{-1}(30.58)\\&\approx 88.13^{\circ}\end{align}
and for the second value of \(x\)
\begin{align}A&= \tan^{-1}(0.67)\\&\approx 33.82^{\circ}\end{align}
Therefore, if the object is to strike the target, the angle of projection must be either \(88.13^{\circ}\) or \(33.82^{\circ}\).
Question 2
An object is projected from \(o\) on the horizontal ground.
If the object’s trajectory passes through the points \(3\vec{i}+6\vec{j}\mbox{ m}\) and \(9\vec{i}+4\vec{j}\mbox{ m}\) relative to \(o\), what is the object’s initial velocity?
\(\vec{u} = 4.35\vec{i} + 12.08\vec{j}\mbox{ m/s}\)
First, let us use an \(x\) component equation with \(s_x=3\) at \(t=t_1\).
\begin{align} a_x=0 && s_x=3 && t=t_1 && u_x =\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}3 &= u_xt_1+\frac{1}{2}(0)t_1^2\\&=u_xt_1\end{align}
\[\,\]
Next, we should use a \(y\) component equation with \(s_y=6\) at \(t=t_1\).
\begin{align} a_y=-9.8 && s_y=6 && t=t_1 && u_y = \mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}6 &= u_yt_1 +\frac{1}{2}(-g)t_1^2\\ &=u_yt_1 – \frac{1}{2} g t_1^2\end{align}
\[\,\]
Let us next use an \(x\) component equation with \(s_x=9\) at \(t=t_2\).
\begin{align} a_x=0 && s_x=9 && t=t_2 && u_x = \mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}9 &= u_xt_2+\frac{1}{2}(0)t_2^2\\&=u_xt_2\end{align}
\[\,\]
Finally, we should use a \(y\) component equation with \(s_y=4\) at \(t=t_2\).
\begin{align} a_y=-9.8 && s_y=4 && t=t_2 && u_y=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}4 = u_yt_2 +\frac{1}{2}(-g)t_2^2\end{align}
\[\,\]
We therefore have the following four equations with four unknowns
\begin{align}3 = u_xt_1\end{align}
\begin{align}6 = u_yt_1 -\frac{1}{2}gt_2^2\end{align}
\begin{align}9 = u_xt_2\end{align}
\begin{align}4 = u_yt_2 -\frac{1}{2}gt_2^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6 = \frac{3u_y}{u_x} -\frac{9g}{2u_x^2}\end{align}
\begin{align}4 = \frac{9u_y}{u_x} -\frac{81g}{2u_x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}18 = \frac{9u_y}{u_x} -\frac{27g}{2u_x^2}\end{align}
\begin{align}4 = \frac{9u_y}{u_x} -\frac{81g}{2u_x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}14 = \frac{27g}{u_x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u_x&=\sqrt{\frac{27g}{14}}\\&\approx 4.35 \mbox{ m/s}\end{align}
Rearranging the following for \(u_y\)
\begin{align}6 = \frac{3u_y}{u_x} -\frac{9g}{2u_x^2}\end{align}
we obtain
\begin{align}u_y &= \frac{u_x}{3}\left(6 + \frac{9g}{2u_x^2}\right)\\&= \frac{4.35}{3}\left(6 + \frac{9(9.8)}{2(4.35^2)}\right)\\&\approx 12.08 \mbox{ m/s}\end{align}
Therefore, the object’s initial velocity is
\begin{align}\vec{u} = 4.35\vec{i} + 12.08\vec{j}\mbox{ m/s}\end{align}
We know \(s_x\) and \(s_y\) at two different times \(t_1\) and \(t_2\) and we wish to find both \(u_x\) and \(u_y\).
Therefore, let us write down four equations.
- an \(x\) component equation with \(s_x=3\) at \(t=t_1\)
- a \(y\) component equation with \(s_y=6\) at \(t=t_1\)
- an \(x\) component equation with \(s_x=9\) at \(t=t_2\)
- a \(y\) component equation with \(s_y=4\) at \(t=t_2\)
First, let us use an \(x\) component equation with \(s_x=3\) at \(t=t_1\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} a_x=0 && s_x=3 \end{align}
and we are looking for \(t=t_1\) and \(u_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}3 &= u_xt_1+\frac{1}{2}(0)t_1^2\\&=u_xt_1\end{align}
Next, we should use a \(y\) component equation with \(s_y=6\) at \(t=t_1\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} a_y=-9.8 && s_y=6 \end{align}
and we are looking for \(t=t_1\) and \(u_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}6 &= u_yt_1 +\frac{1}{2}(-g)t_1^2\\ &=u_yt_1 – \frac{1}{2} g t_1^2\end{align}
Let us next use an \(x\) component equation with \(s_x=9\) at \(t=t_2\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} a_x=0 && s_x=9 \end{align}
and we are looking for \(t=t_2\) and \(u_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}9 &= u_xt_2+\frac{1}{2}(0)t_2^2\\&=u_xt_2\end{align}
Finally, we should use a \(y\) component equation with \(s_y=4\) at \(t=t_2\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} a_y=-9.8 && s_y=4 \end{align}
and we are looking for \(t=t_2\) and \(u_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}4 = u_yt_2 +\frac{1}{2}(-g)t_2^2\end{align}
We therefore have the following four equations with four unknowns
\begin{align}3 = u_xt_1\end{align}
\begin{align}6 = u_yt_1 -\frac{1}{2}gt_2^2\end{align}
\begin{align}9 = u_xt_2\end{align}
\begin{align}4 = u_yt_2 -\frac{1}{2}gt_2^2\end{align}
As we are looking for \(u_x\) and \(u_y\), we should try to eliminate \(t_1\) and \(t_2\) from these equations. This can be done by inserting \(t_1=\displaystyle\frac{3}{u_x}\) from the first equation into the second equation and \(t_2=\displaystyle\frac{9}{u_x}\) from the third equation into the fourth equation, leaving us with
\begin{align}6 = \frac{3u_y}{u_x} -\frac{9g}{2u_x^2}\end{align}
\begin{align}4 = \frac{9u_y}{u_x} -\frac{81g}{2u_x^2}\end{align}
We can now eliminate \(u_y\) by multiplying the first equation by \(3\)
\begin{align}18 = \frac{9u_y}{u_x} -\frac{27g}{2u_x^2}\end{align}
\begin{align}4 = \frac{9u_y}{u_x} -\frac{81g}{2u_x^2}\end{align}
and subtracting the second equation from the first equation, giving
\begin{align}14 = \frac{27g}{u_x^2}\end{align}
Solving for \(u_x\), we obtain
\begin{align}u_x&=\sqrt{\frac{27g}{14}}\\&\approx 4.35 \mbox{ m/s}\end{align}
We can now use this to find \(u_y\) by rearranging e.g.
\begin{align}6 = \frac{3u_y}{u_x} -\frac{9g}{2u_x^2}\end{align}
for \(u_y\) to obtain
\begin{align}u_y &= \frac{u_x}{3}\left(6 + \frac{9g}{2u_x^2}\right)\\&= \frac{4.35}{3}\left(6 + \frac{9(9.8)}{2(4.35^2)}\right)\\&\approx 12.08 \mbox{ m/s}\end{align}
Therefore, the object’s initial velocity is
\begin{align}\vec{u} = 4.35\vec{i} + 12.08\vec{j}\mbox{ m/s}\end{align}
Question 3
A particle is projected from \(o\) on the horizontal ground with an initial speed of \(26\mbox{ m s}^{-1}\) at an angle \(\alpha\), where \(\tan \alpha = \displaystyle\frac{5}{12}\).
(a) Write the initial velocity vector in component form.
(b) What is the object’s displacement after \(1\) second?
(c) What angle relative to the ground is the object moving through the air at after \(1\) second?
(a) \(\vec{u}=24\vec{i} + 10\vec{j}\mbox{ m/s}\)
(b) \(\vec{s} = 24\vec{i}+5.1 \vec{j}\mbox{ m}\)
(c) \(0.48^{\circ}\)
(a) The hypotenuse of the corresponding triangle is
\begin{align}H &= \sqrt{5^2+12^2} \\&= 13\end{align}
and therefore
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
\begin{align}\downarrow\end{align}
\[\begin{align}|\vec{u}_x| &= |\vec{u}| \cos A \\& = 26 \left(\frac{12}{13}\right) \\&= 24\end{align}\]
\[\begin{align}|\vec{u}_y| &= |\vec{u}| \sin A \\&= 26 \left(\frac{5}{13}\right) \\&= 10\end{align}\]
\begin{align}\downarrow\end{align}
\begin{align}u_x = 24 && u_y = 10\end{align}
\begin{align}\downarrow\end{align}
\[\vec{u}=24\vec{i} + 10\vec{j}\mbox{ m/s}\]
(b)
First, the \(x\) component of the displacement.
\begin{align} u_x = 24 && a_x=0 && t=1 && s_x = \mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}s_x&=(24)(1)+\frac{1}{2}(0)(1^2)\\&=24 \mbox{ m}\end{align}
\[\,\]
Next, the \(y\) component of the displacement.
\begin{align} u_y = 10 && a_y=-9.8 && t=1 && s_y=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}s_y&=(10)(1)+\frac{1}{2}(-9.8)(1^2)\\&=5.1 \mbox{ m}\end{align}
\[\,\]
Therefore, the object’s displacement from \(o\) after \(1\) second is
\begin{align}\vec{s} = 24\vec{i}+5.1 \vec{j}\mbox{ m}\end{align}
(c)
First, the \(x\) component of the velocity.
\begin{align} u_x = 24 && a_x=0 && t=1 && s_x= 24 && v_x = \mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[v_x=u_x+a_xt\]
\begin{align}\downarrow\end{align}
\begin{align}v_x&=24+(0)(1)\\&=24 \mbox{ m/s}\end{align}
\[\,\]
Next, the \(y\) component of the velocity.
\begin{align} u_y = 10 && a_y=-9.8 && t=1 && s_y= 5.1 && v_y=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[v_y=u_y+a_yt\]
\begin{align}\downarrow\end{align}
\begin{align}v_y&=10+(-9.8)(1)\\&=0.2 \mbox{ m/s}\end{align}
\[\,\]
Therefore, the velocity of the object after \(1\) second is
\begin{align}\vec{v}=24\vec{i}+0.2\vec{j}\end{align}
and hence
\begin{align}\theta &= \tan^{-1}\left(\frac{0.2}{24}\right)\\&\approx 0.48^{\circ}\end{align}
(a)
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\alpha}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\alpha\) given in polar form is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{5}{12}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{u_x}| = |\vec{u}|\cos A} \) and \(\boldsymbol{|\vec{u_y}| = |\vec{u}|\sin A} \).
Let’s draw the following separate triangle which correctly includes this same angle \(A\).
Let us now calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{5^2+12^2} \\&= 13\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
Using these, we can directly calculate our components as
\[\begin{align}|\vec{u}_x| &= |\vec{u}| \cos A \\& = 26 \left(\frac{12}{13}\right) \\&= 24\end{align}\]
\[\begin{align}|\vec{u}_y| &= |\vec{u}| \sin A \\&= 26 \left(\frac{5}{13}\right) \\&= 10\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{u_x}\) and \(\boldsymbol{u_y}\) using \(\boldsymbol{|\vec{u_x}|}\) and \(\boldsymbol{|\vec{u_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{u_x}\) is in the same direction as \(\vec{i}\) and \(u_y\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}u_x = 24 && u_y = 10\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{u} =u_x \vec{i} + u_y \vec{j}}\).
\[\vec{u}=24\vec{i} + 10\vec{j}\mbox{ m/s}\]
(b) We are being asked to find \(\vec{s} = s_x\vec{i} + s_y\vec{j}\) when \(t=1\). We therefore need to calculate both \(s_x\) and \(s_y\).
Let us start by finding \(s_x\). We therefore need to use an \(x\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 24 && a_x=0 && t=1 \end{align}
and we are looking for \(s_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}s_x&=(24)(1)+\frac{1}{2}(0)(1^2)\\&=24 \mbox{ m}\end{align}
Let us next find \(s_y\). We therefore need to use an \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 10 && a_y=-9.8 && t=1 \end{align}
and we are looking for \(s_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(x\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}s_y&=(10)(1)+\frac{1}{2}(-9.8)(1^2)\\&=5.1 \mbox{ m}\end{align}
Therefore, the object’s displacement from \(o\) after \(1\) second is
\begin{align}\vec{s} = 24\vec{i}+5.1 \vec{j}\mbox{ m}\end{align}
(c) As the velocity vector points along the particle’s direction, we are therefore being asked to find the angle of the velocity vector after \(1\) second.
Hence, we first need to find \(\vec{v} = v_x\vec{i} + v_y\vec{j}\) when \(t=1\). We therefore need to calculate both \(v_x\) and \(v_y\).
Let us start by finding \(v_x\). We therefore need to use an \(x\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 24 && a_x=0 && t=1 && s_x= 24 \end{align}
and we are looking for \(v_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we know four of the five \(x\) component SUVAT variables, we can use any SUVAT equation that includes \(v_x\). I shall randomly choose
\[v_x=u_x+a_xt\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}v_x&=24+(0)(1)\\&=24 \mbox{ m/s}\end{align}
Let us next find \(v_y\). We therefore need to use an \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 10 && a_y=-9.8 && t=1 && s_y= 5.1 \end{align}
and we are looking for \(v_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we know four of the five \(y\) component SUVAT variables, we can use any SUVAT equation that includes \(v_y\). We shall randomly choose
\[v_y=u_y+a_yt\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}v_y&=10+(-9.8)(1)\\&=0.2 \mbox{ m/s}\end{align}
Therefore, the velocity of the object after \(1\) second is
\begin{align}\vec{v}=24\vec{i}+0.2\vec{j}\end{align}
The angle of the object at this time is angle of this vector relative to the horizonal ground, i.e.
\begin{align}\theta &= \tan^{-1}\left(\frac{v_y}{v_x}\right)\\&=\tan^{-1}\left(\frac{0.2}{24}\right)\\&\approx 0.48^{\circ}\end{align}
Question 4
A particle is projected from \(o\) with a speed of \(15\vec{i}+10\vec{j}\mbox{ m/s}\).
(a) What is the height of the particle when its horizontal displacement has a magnitude that is \(80\%\) of the object’s range?
(b) At what other time is the object at this height?
(a) \(3.28\mbox{ m}\)
(b) \(0.61\mbox{ s}\)
(a)
Time of Flight
\begin{align} u_y = 10 && a_y=-9.8 && s_y=0 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}0=10t +\frac{1}{2}(-9.8)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t\left(t-\frac{2(10)}{9.8}\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{2(10)}{9.8}\\&\approx 2.04 \mbox{ s} \end{align}
\[\,\]
Range
\begin{align} u_x = 15 && a_x=0 && t=2.04 && s_x=R\end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}R &= (15)(2.04)+\frac{1}{2}(0)(2.04^2)\\&\approx 30.6 \mbox{ m}\end{align}
and therefore
\begin{align}s_x&=0.8R\\&=0.8(30.6)\\&=24.48 \mbox{ m}\end{align}
\[\,\]
We shall first use this to find the time.
\begin{align} u_x = 15 && a_x=0 && s_x=24.48 && t=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}24.48=15t +\frac{1}{2}(0)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{24.48}{15}\\&\approx 1.63 \mbox{s}\end{align}
\[\,\]
Now, we will use this time to find the height.
\begin{align} u_y = 10 && a_y=-9.8 && t=1.63 && s_y=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}s_y&= (10)(1.63)+\frac{1}{2}(-9.8)(1.63^2)\\&\approx 3.28 \mbox{ m} \end{align}
(b)
\begin{align} u_y = 10 && a_y=-9.8 && s_y=3.28 && t=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}3.28 = 10t+\frac{1}{2}(-9.8)t^2 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}4.9t^2-10t+3.28=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{10 \pm \sqrt{(-10)^2-4(4.9)(3.28)} }{2(4.9)}\\&\approx 1.02 \pm 0.61 \end{align}
Therefore, this object is at this height after \(0.61\) and \(1.63\) seconds. As expected, the larger time agrees with the time we found in part (b). The other time the object is at this height is therefore \(0.61\) seconds.
(a) We are being asked to find \(s_x\) when \(s_y=0.8R\). Therefore, let us first find the range \(R\) before using it in this condition.
The range is given by the horizontal displacement coefficient \(s_x\) when \(s_y=0\).
As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 10 && a_y=-9.8 && s_y=0 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=10t +\frac{1}{2}(-9.8)t^2\end{align}
Factorising, this gives
\begin{align}t\left(t-\frac{2(10)}{9.8}\right)=0\end{align}
Therefore, either \(t=0\) or
\begin{align}t&=\frac{2(10)}{9.8}\\&\approx 2.04 \mbox{ s} \end{align}
The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is \(t=2.04\mbox{ s}\).
We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 15 && a_x=0 && t=2.04 \end{align}
and we are looking for \(s_x=R\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}R &= (15)(2.04)+\frac{1}{2}(0)(2.04^2)\\&\approx 30.6 \mbox{ m}\end{align}
Hence, returning to the original question, we are therefore being asked to find \(s_y\) when
\begin{align}s_x&=0.8R\\&=0.8(30.6)\\&=24.48 \mbox{ m}\end{align}
As this statement defines an \(y\) component based on the condition of an \(x\) component, this suggests that we should first find \(t\) using an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 15 && a_x=0 && s_x=24.48 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}24.48=15t +\frac{1}{2}(0)t^2\end{align}
Rearranging for \(t\), this gives
\begin{align}t&=\frac{24.48}{15}\\&\approx 1.63 \mbox{s}\end{align}
We now wish to find what \(s_y\) is at this time. We therefore need to use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 10 && a_y=-9.8 && t=1.63 \end{align}
and we are looking for \(s_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}s_y&= (10)(1.63)+\frac{1}{2}(-9.8)(1.63^2)\\&\approx 3.28 \mbox{ m} \end{align}
(b) The object will be at this height at two times – on the way up and on the way down. In part (b), as the horizontal displacement was \(80\%\) of the range, we therefore found the time on the way down. We now wish to find the time that the object will be at this height instead on the way up.
We will therefore find the time \(t\) when \(s_y=3.28\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 10 && a_y=-9.8 && s_y=3.28 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}3.28 = 10t+\frac{1}{2}(-9.8)t^2 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}4.9t^2-10t+3.28=0\end{align}
This can be solved using the quadratic formula to give
\begin{align}t&=\frac{10 \pm \sqrt{(-10)^2-4(4.9)(3.28)} }{2(4.9)}\\&\approx 1.02 \pm 0.61 \end{align}
Therefore, this object is at this height after \(0.61\) and \(1.63\) seconds. As expected, the larger time agrees with the time we found in part (b). The other time the object is at this height is therefore \(0.61\) seconds.
Question 5
What angle of projection will result in the maximum height equalling the range of a projected particle?
\(75.96^{\circ}\)
Maximum Height
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 && s_y=H\end{align}
\begin{align}\downarrow\end{align}
\[v_y^2 = u_y^2+2a_ys_y\]
\begin{align}\downarrow\end{align}
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
\begin{align}\downarrow\end{align}
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
\[\,\]
Time of Flight
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
\[\,\]
Range
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} && s_x=R \end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\\&=\frac{u^2(2 \sin A\cos A)}{g}\\&=\frac{u^2 \sin 2 A}{g}\end{align}
\[\,\]
Setting these two equations equal to each other then gives
\begin{align}\frac{u^2\sin^2A}{2g}=\frac{u^2\sin 2A}{g} \end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\sin^2 A}{2} = \sin 2A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\sin^2 A}{2} = 2\sin A \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin A = 4 \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan A = 4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&= \tan^{-1} (4)\\&\approx 75.96^{\circ}\end{align}
We wish to compare the maximum height formula and the range formula. Let us derive each one first, starting with the maximum height formula.
The maximum height is equal to the value of \(s_y\) when \(v_y=0\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are therefore given the following
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 \end{align}
and we are looking for \(s_y=H\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely
\[v_y^2 = u_y^2+2a_ys_y\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
Rearranging, this gives
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
Next, let us derive the range formula. We wish to find the horizontal displacement coefficient \(s_x\) when \(s_y=0\).
As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are therefore given the following
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
Factorising, this gives
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
Therefore, either \(t=0\) or
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
as required.
We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are therefore given the following
\begin{align} u_x = u \cos A && a_x=0 && t=\frac{2u \sin A}{g} \end{align}
and we are looking for \(s_x=R\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}R &=(u \cos A)\left(\frac{2u\sin A}{g}\right)+\frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&=\frac{2u^2 \sin A\cos \theta}{g}\end{align}
Inserting the following identity
\begin{align}\sin 2A = 2 \sin A \cos A\end{align}
then gives
\begin{align}R &=\frac{2u^2 \sin A\cos A}{g}\\&=\frac{u^2(2 \sin A\cos A)}{g}\\&=\frac{u^2 \sin 2 A}{g}\end{align}
\[\,\]
Setting these two equations equal to each other then gives
\begin{align}\frac{u^2\sin^2A}{2g}=\frac{u^2\sin 2A}{g} \end{align}
Cancelling like terms reduces this to
\begin{align}\frac{\sin^2 A}{2} = \sin 2A\end{align}
How do we solves this? Well let’s first undo the trigonometric identity we used for the range so all of the term have only \(A\) as the argument rather than \(2A\).
\begin{align}\frac{\sin^2 A}{2} = 2\sin A \cos A\end{align}
We can now cancel a \(\sin A\) term from both sides
\begin{align}\sin A = 4 \cos A\end{align}
Rearranging, we then obtain
\begin{align}\tan A = 4\end{align}
and therefore
\begin{align}A&= \tan^{-1} (4)\\&\approx 75.96^{\circ}\end{align}
Question 6
An object is projected horizontally from a cliff with a speed of \(20\mbox{ m s}^{-1}\). If the cliff is \(100\) metres high:
(a) how far from the base of the cliff does the object hit the ocean?
(b) what angle relative to the ocean does the object hit the ocean?
(a) \(90.40\mbox{ m}\)
(b) \(65.69^{\circ}\)
(a)
First, we will find the time taken to hit the ocean.
\begin{align} u_y = 0 && a_y=-9.8 && s_y=-100 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}-100=(0)t+\frac{1}{2}(-9.8)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\sqrt{\frac{(-100)(2)}{-9.8}}\\&\approx 4.52 \mbox{ s}\end{align}
\[\,\]
We will now use this time to find this distance.
\begin{align} u_x = 20 && a_x=0 && t=4.52 && s_x = \mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}s_x&=(20)(4.52)+\frac{1}{2}(0)(4.52^2)\\&=90.40 \mbox{ m}\end{align}
(b)
\(v_x=u_x=20\).
The \(y\) component is instead found as follows.
\begin{align} u_y = 0 && a_y=-9.8 && s_y= -100 && v_y=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[v_y^2=u_y^2+2a_ys_y\]
\begin{align}\downarrow\end{align}
\begin{align}v_y^2 = 0^2+2(-9.8)(-100)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=-\sqrt{2(-9.8)(-100)}\\&\approx -44.27 \mbox{ m/s}\end{align}
Therefore, the velocity of the object when the object hits the ocean is
\begin{align}\vec{v}=20\vec{i}-44.27\vec{j}\end{align}
Hence, the angle of the object at this time is
\begin{align}\theta &= \tan^{-1}\left(\frac{44.27}{20}\right)\\&\approx 65.69^{\circ}\end{align}
(a) The initial velocity coefficients are
\begin{align}u_x = 20 && u_y=0\end{align}
We are being asked to find \(s_x\) when \(s_y=-100\).
As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 0 && a_y=-9.8 && s_y=-100 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}-100=(0)t+\frac{1}{2}(-9.8)t^2\end{align}
Rearranging for \(t\), this gives
\begin{align}t&=\sqrt{\frac{(-100)(2)}{-9.8}}\\&\approx 4.52 \mbox{ s}\end{align}
Now that we know \(t\), we now know enough \(x\) component quantities to find \(s_x\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 20 && a_x=0 && t=4.52\end{align}
and we are looking for \(s_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}s_x&=(20)(4.52)+\frac{1}{2}(0)(4.52^2)\\&=90.40 \mbox{ m}\end{align}
(b) As the velocity vector points along the particle’s direction, we are therefore being asked to find the angle of the velocity vector when the object hits the ground, i.e. when \(\vec{s} = 90.40\vec{i} -100\vec{j}\).
Hence, we first need to find \(\vec{v} = v_x\vec{i} + v_y\vec{j}\) at this displacement. We therefore need to calculate both \(v_x\) and \(v_y\).
As there is no acceleration in the \(x\) direction, we can state immediately that \(v_x=u_x=20\).
To find \(v_y\), we need to use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 0 && a_y=-9.8 && s_y= -100 \end{align}
and we are looking for \(v_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1,
As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely
\[v_y^2=u_y^2+2a_ys_y\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}v_y^2 = 0^2+2(-9.8)(-100)\end{align}
We will take the negative root as we expect the velocity vector to be pointing downwards.
\begin{align}v_y&=-\sqrt{2(-9.8)(-100)}\\&\approx -44.27 \mbox{ m/s}\end{align}
Therefore, the velocity of the object when the object hits the ocean is
\begin{align}\vec{v}=20\vec{i}-44.27\vec{j}\end{align}
Hence, the angle of the object at this time is
\begin{align}\theta &= \tan^{-1}\left(\frac{44.27}{20}\right)\\&\approx 65.69^{\circ}\end{align}
Question 7
An object is projected from \(o\) with a speed \(u=15\mbox{ m/s}\) at an angle \(A\) relative to the horizontal ground.
(a) If the projectile is to hit a target that is located at \(3\vec{i}-2\vec{j}\mbox{ m}\) from \(o\), what are the two possible values of \(A\)?
(b) How long does it take the object to hit the target in both cases?
(a) \(1.15^{\circ}\) and \(56.66^{\circ}\)
(b) \(0.20\mbox{ s}\) and \(0.36\mbox{ s}\)
(a)
We will first use an \(x\) component SUVAT equation.
\begin{align} u_x = 15\cos A && a_x=0 && s_x=3 && t=\mathord{?}\end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}3=15\cos A t + \frac{1}{2}(0)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t = \frac{1}{5\cos A}\end{align}
\[\,\]
Next, we shall use a \(y\) component SUVAT equation.
\begin{align} u_y = 15\sin A && a_y=-9.8 && s_y=-2 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}-2=15\sin A t -\frac{1}{2}(9.8)t^2\end{align}
\[\,\]
If we now sub our first equation for \(t\) into this equation, we get
\begin{align}-2&=15\sin A\left(\frac{1}{5\cos A}\right) -\frac{1}{2}(9.8)\left(\frac{1}{5 \cos A}\right)^2\\&= 3\tan A-\frac{1.96}{\cos^2 A} \end{align}
Inserting the following trigonometric identity
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
we then obtain
\begin{align}-2 = 3 \tan A- 1.96(1+\tan^2 A)\end{align}
Setting \(x=\tan A\), this becomes
\begin{align}-2=3x-1.96(1+x^2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1.96 x^2-3x-0.04=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{3 \pm \sqrt{(-3)^2-4(1.96)(0.04)}}{2(1.96)}\\ &\approx 0.77 \pm 0.75\end{align}
and therefore \(x=0.02\) or \(x=1.52\). Hence
\begin{align}A&= \tan^{-1}(0.02)\\&\approx 1.15^{\circ}\end{align}
and
\begin{align}A&= \tan^{-1}(1.52)\\&\approx 56.66^{\circ}\end{align}
(b)
\begin{align}t = \frac{1}{5\cos A}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t &= \frac{1}{5\cos 1.15}\\&\approx 0.20 \mbox{ s} \end{align}
and
\begin{align}t &= \frac{1}{5\cos 56.66}\\&\approx 0.36 \mbox{ s} \end{align}
(a) To answer this, we should find the time taken for the object to reach \(s_x = 3\) and \(s_y=-2\) separately before equating those times and solving for \(A\).
To start, we first wish to find \(t\) when \(s_x=3\). We should therefore use an \(x\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 15\cos A && a_x=0 && s_x=3 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}3=15\cos A t + \frac{1}{2}(0)t^2\end{align}
Rearranging, this gives
\begin{align}t = \frac{1}{5\cos A}\end{align}
Next, we should find \(t\) when \(s_y=-2\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 15\sin A && a_y=-9.8 && s_y=-2 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}-2=15\sin A t -\frac{1}{2}(9.8)t^2\end{align}
If we now sub our first equation for \(t\) into this equation, we get
\begin{align}-2&=15\sin A\left(\frac{1}{5\cos A}\right) -\frac{1}{2}(9.8)\left(\frac{1}{5 \cos A}\right)^2\\&= 3\tan A-\frac{1.96}{\cos^2 A} \end{align}
Inserting the following trigonometric identity
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
we then obtain
\begin{align}-2 = 3 \tan A- 1.96(1+\tan^2 A)\end{align}
Now, let us temporarily set \(x=\tan A\), reducing this equation to
\begin{align}-2=3x-1.96(1+x^2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1.96 x^2-3x-0.04=0\end{align}
This can be solved using the quadratic formula to obtain
\begin{align}x&=\frac{3 \pm \sqrt{(-3)^2-4(1.96)(0.04)}}{2(1.96)}\\ &\approx 0.77 \pm 0.75\end{align}
and therefore \(x=0.02\) or \(x=1.52\). Inverting this, we therefore obtain for the first value of \(x\)
\begin{align}A&= \tan^{-1}(0.02)\\&\approx 1.15^{\circ}\end{align}
and for the second value of \(x\)
\begin{align}A&= \tan^{-1}(1.52)\\&\approx 56.66^{\circ}\end{align}
Therefore, if the object is to strike the target, the angle of projection must be either \(1.15^{\circ}\) or \(56.66^{\circ}\).
(b) We can use either formula that we derived for \(t\). We shall randomly choose
\begin{align}t = \frac{1}{5\cos A}\end{align}
For the first the first case of \(A=1.15^{\circ}\), we obtain
\begin{align}t &= \frac{1}{5\cos 1.15^{\circ}}\\&\approx 0.20 \mbox{ s} \end{align}
and for the first the second case of \(A=56.66^{\circ}\), we obtain
\begin{align}t &= \frac{1}{5\cos 56.66^{\circ}}\\&\approx 0.36 \mbox{ s} \end{align}
Question 8
A particle is projected from \(o\). The maximum height of the particle is \(10\) metres and it lands \(100\) metres from \(o\).
What was the initial speed of the object?
\(37.70\mbox{ m/s}\)
Maximum Height
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 && s_y=H\end{align}
\begin{align}\downarrow\end{align}
\[v_y^2 = u_y^2+2a_ys_y\]
\begin{align}\downarrow\end{align}
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
\begin{align}\downarrow\end{align}
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
\[\,\]
Time of Flight
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
\[\,\]
Range
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} && s_x=R \end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
Inserting our values into both equations, we obtain
\begin{align}10 = \frac{u^2\sin^2A}{2g}\end{align}
\begin{align}100 = \frac{2u^2\sin A \cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}20g = u^2\sin^2A\end{align}
\begin{align}100g = 2u^2\sin A \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}100g = 5u^2\sin^2A\end{align}
\begin{align}100g = 2u^2\sin A \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5u^2\sin^2A = 2u^2\sin A \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5 \sin A = 2 \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan A = \frac{2}{5}\end{align}
Hence, the angle of projection is
\begin{align}A&= \tan^{-1}\left(\frac{2}{5}\right)\\&\approx 21.80^{\circ} \end{align}
Inserting this into
\begin{align}10 = \frac{u^2\sin^2A}{2g}\end{align}
gives
\begin{align}u&=\sqrt{\frac{(10(2g)}{\sin^2(21.80)}}\\&\approx 37.70 \mbox{ m/s}\end{align}
We are told both the maximum height and the range. Let us therefore derive each formula first, starting with the maximum height formula.
The maximum height is equal to the value of \(s_y\) when \(v_y=0\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 \end{align}
and we are looking for \(s_y=H\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely
\[v_y^2 = u_y^2+2a_ys_y\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
Rearranging, this gives
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
Next, let us derive the range formula. We wish to find the horizontal displacement coefficient \(s_x\) when \(s_y=0\).
As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are therefore given the following
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
Factorising, this gives
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
Therefore, either \(t=0\) or
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
as required.
We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are therefore given the following
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} \end{align}
and we are looking for \(s_x=R\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s),
Subbing in our known values, we obtain
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
If we now sub in the given values for both the maximum height and range, we then obtain
\begin{align}10 = \frac{u^2\sin^2A}{2g}\end{align}
\begin{align}100 = \frac{2u^2\sin A \cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}20g = u^2\sin^2A\end{align}
\begin{align}100g = 2u^2\sin A \cos A\end{align}
We can easily make the left hand side of both equations the same by multiplying the first equation by \(5\), resulting in
\begin{align}100g = 5u^2\sin^2A\end{align}
\begin{align}100g = 2u^2\sin A \cos A\end{align}
We can now set the right hand sides equal to each other, giving
\begin{align}5u^2\sin^2A = 2u^2\sin A \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5 \sin A = 2 \cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan A = \frac{2}{5}\end{align}
Hence, the angle of projection is
\begin{align}A&= \tan^{-1}\left(\frac{2}{5}\right)\\&\approx 21.80^{\circ} \end{align}
We can now use this angle in either of our equations above to find \(u\). We shall randomly choose
\begin{align}10 = \frac{u^2\sin^2A}{2g}\end{align}
Rearranging, this gives
\begin{align}u&=\sqrt{\frac{(10(2g)}{\sin^2(21.80)}}\\&\approx 37.70 \mbox{ m/s}\end{align}
Question 9
A particle is projected from \(o\).
If the range is five time the maximum height, what was the angle of projection?
\(38.66^{\circ}\)
Maximum Height
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 && s_y=H\end{align}
\begin{align}\downarrow\end{align}
\[v_y^2 = u_y^2+2a_ys_y\]
\begin{align}\downarrow\end{align}
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
\begin{align}\downarrow\end{align}
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
\[\,\]
Time of Flight
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
\[\,\]
Range
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} && s_x=R \end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
\[\,\]
These two equations then combine to give
\begin{align}\frac{2 u^2\sin A \cos A}{g}= 5\left(\frac{u^2\sin^2A}{2g}\right) \end{align}
\begin{align}\downarrow\end{align}
\begin{align}2 \cos A = \frac{5}{2} \sin A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan A = \frac{4}{5}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\tan^{-1}\left(\frac{4}{5}\right)\\&\approx 38.66^{\circ}\end{align}
We wish to compare the maximum height formula and the range formula. Let us derive each one first, starting with the maximum height formula.
The maximum height is equal to the value of \(s_y\) when \(v_y=0\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = u\sin A&& a_y=-g && v_y=0 \end{align}
and we are looking for \(s_y=H\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
\[\,\]
As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely
\[v_y^2 = u_y^2+2a_ys_y\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
\[\,\]
Subbing in our known values, we obtain
\begin{align}0^2 = u^2 \sin^2 A+ 2(-g)H\end{align}
Rearranging, this gives
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
Next, let us derive the range formula. We wish to find the horizontal displacement coefficient \(s_x\) when \(s_y=0\).
As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
Factorising, this gives
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
Therefore, either \(t=0\) or
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
as required.
We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
With the initial direction of motion as stated, we are therefore given the following
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} \end{align}
and we are looking for \(s_x=R\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
As we are told the range is five times the maximum height, these two equations then combine to give
\begin{align}\frac{2 u^2\sin A \cos A}{g}= 5\left(\frac{u^2\sin^2A}{2g}\right) \end{align}
\begin{align}\downarrow\end{align}
\begin{align}2 \cos A = \frac{5}{2} \sin A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan A = \frac{4}{5}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\tan^{-1}\left(\frac{4}{5}\right)\\&\approx 38.66^{\circ}\end{align}
Question 10
Object \(A\) is moving horizontally with a constant speed of \(15\mbox{ m s}^{-1}\) at a height of \(20\) metres from the ground.
As that object passes a point \(P\), object \(B\) is launched directly below \(P\) from the ground with a speed of \(25\mbox{ m s}^{-1}\) at an angle \(\alpha\) relative the ground.
If the objects collide as object \(B\) is travelling upwards:
(a) what is the value of \(\alpha\)?
(b) how long is object \(B\) in the air until the collision occurs?
(a) \(53.13^{\circ}\)
(b) \(1.75\mbox{ s}\)
(a)
\begin{align}u_x &= 25 \cos\alpha \\&=15 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}\alpha & = \cos^{-1}\left(\frac{15}{25}\right)\\&\approx 53.13^{\circ} \end{align}
(b)
\begin{align} u_y = 25\sin 53.13 && a_y=-9.8 && s_y=20 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}20=(25 \sin 53.13^{\circ})t+\frac{1}{2}(-9.8)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4.9t^2-20t+20=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{20 \pm \sqrt{(-20)^2-4(4.9)(20)}}{2(4.9)}\\&\approx 2.04 \pm 0.29\end{align}
Therefore, the time taken for object \(B\) to be at a height of \(20\) metres is \(1.75\) seconds (on the way up) or \(2.33\) seconds (on the way down).
After the objects collides when object \(B\) is on the way up, they won’t again collide when object \(B\) is on the way down. Therefore, the correct time is \(1.75\) seconds.
(a) This question may seem like it’s quite complicated and messy, but it actually isn’t! Below is a diagram of the situation.
The only way these two objects can collide is if both objects have the same horizontal component initial velocity. Otherwise, object \(B\) will not remain directly below object \(A\) as it rises and there will be no collision.
Hence, for object \(B\), we must have that
\begin{align}u_x &= 25 \cos\alpha \\&=15 \end{align}
Solving for \(\alpha\), this gives
\begin{align}\alpha & = \cos^{-1}\left(\frac{15}{25}\right)\\&\approx 53.13^{\circ} \end{align}
(b) This also might seem complicated but again it isn’t! We are simply being asked to find \(t\) for object \(B\) when \(s_y=20\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 25\sin 53.13 && a_y=-9.8 && s_y=20 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}20=(25 \sin 53.13^{\circ})t+\frac{1}{2}(-9.8)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4.9t^2-20t+20=0\end{align}
This can be solved using the quadratic formula to give
\begin{align}t&=\frac{20 \pm \sqrt{(-20)^2-4(4.9)(20)}}{2(4.9)}\\&\approx 2.04 \pm 0.29\end{align}
Therefore, the time taken for object \(B\) to be at a height of \(20\) metres is \(1.75\) seconds (on the way up) or \(2.33\) seconds (on the way down).
Of course, after the objects collide when object \(B\) is on the way up, they won’t again collide when object \(B\) is on the way down!
Therefore, the correct time is \(1.75\) seconds.
Question 11
An object is projected from the ground at \(o\) and lands \(20\) metres from \(o\) after \(5\) seconds.
(a) What was the object’s initial velocity?
During its flight, it just passes over a fence located \(5\) metres from \(o\).
(b) What is the height of the fence?
(c) How long did it take the object to reach the fence?
(a) \(\vec{u}= 4\vec{i} + 24.5\vec{j}\mbox{ m/s}\)
(b) \(22.97\mbox{ m}\)
(c) \(1.25\mbox{ s}\)
(a)
Time of Flight
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
\[\,\]
Range
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} && s_x=R \end{align}
\begin{align}\downarrow\end{align}
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
\[\,\]
We therefore have the following two equations
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}20=\frac{8 u \sin A}{g}\end{align}
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{8 u \sin A}{g}=\frac{2u^2 \sin A\cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u \cos A = 4\end{align}
Inserting \(u = \displaystyle\frac{4}{\cos A}\) into
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
gives
\begin{align}5=\frac{2 \left(\frac{4}{\cos a}\right)\sin A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5=\frac{8 \tan A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\tan^{-1}\left(\frac{5g}{8}\right)\\&\approx 80.73^{\circ}\end{align}
Using
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
we obtain
\begin{align}u &= \frac{5g}{2 \sin A}\\&=\frac{5(9.8)}{2 \sin 80.73^{\circ}} \\&\approx 24.82 \mbox{ m/s}\end{align}
Therefore
\begin{align}\vec{u} &= 24.82 \cos 80.73\vec{i} + 24.82 \sin 80.73\vec{j}\\&= 4\vec{i} + 24.5\vec{j}\mbox{ m/s} \end{align}
(b)
\begin{align} u_x =4 && a_x=0 && s_x=5 && t=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}5 &= 4t + \frac{1}{2}(0)t^2\end{align}
and therefore \(t=1.25\mbox{ s}\).
\[\,\]
\begin{align} u_y = 24.5 && a_y=-9.8 && t=1.25 && s_y=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}s_y&=24.5(1.25)+\frac{1}{2}(-9.8)(1.25^2)\\&\approx 22.97 \mbox{ m}\end{align}
(c) As we determined in part (b), it takes the object \(1.25\) seconds to reach the fence.
(a) We are told both the time of flight and the range of the object. Let us therefore derive the formulae for each of these, beginning with the time of flight.
We wish to find the time \(t\) when \(s_y=0\). We should therefore use a \(y\) component SUVAT equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = u\sin A && a_y=-g && s_y=0 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=u \sin A t +\frac{1}{2}(-g)t^2\end{align}
Factorising, this gives
\begin{align}t\left(t-\frac{2 u \sin A}{g}\right)=0\end{align}
Therefore, either \(t=0\) or
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
Inserting the given time of flight, this becomes
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
\[\,\]
Next, let us derive the formula for the range.
We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = u\cos A && a_x=0 && t=\frac{2u \sin A}{g} \end{align}
and we are looking for \(s_x=R\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x=u_xt+\frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}R &= u \cos A\left(\frac{2u\sin A}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin A}{g}\right)^2\\&= \frac{2 u^2 \sin A \cos A}{g}\end{align}
Inserting the given value therefore gives
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
We therefore have the following two equations
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
We can multiply the first equation by \(4\)
\begin{align}20=\frac{8 u \sin A}{g}\end{align}
\begin{align}20=\frac{2u^2 \sin A\cos A}{g}\end{align}
and thereby set these equations equal to each other.
\begin{align}\frac{8 u \sin A}{g}=\frac{2u^2 \sin A\cos A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u \cos A = 4\end{align}
We can now sub in that \(u = \displaystyle\frac{4}{\cos A}\) into either of our original equations, e.g.
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
to obtain
\begin{align}5=\frac{2 \left(\frac{4}{\cos a}\right) \sin A}{g}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5=\frac{8 \tan A}{g}\end{align}
Solving for \(A\), we obtain
\begin{align}A&=\tan^{-1}\left(\frac{5g}{8}\right)\\&\approx 80.73^{\circ}\end{align}
To find \(u\), we can now insert this into any equation above that contains both \(u\) and \(A\), e.g.
\begin{align}5=\frac{2 u \sin A}{g}\end{align}
to obtain
\begin{align}u &= \frac{5g}{2 \sin A}\\&=\frac{5(9.8)}{2 \sin 80.73^{\circ}} \\&\approx 24.82 \mbox{ m/s}\end{align}
Hence, the object’s initial velocity is
\begin{align}\vec{u} &= 24.82 \cos 80.73\vec{i} + 24.82 \sin 80.73\vec{j}\\&= 4\vec{i} + 24.5\vec{j}\mbox{ m/s}\end{align}
(b) We are being asked to find \(s_y\) when \(s_x=5\). As this statement defines an x component based on the condition of a y component, this suggests that we should first find \(t\) using an \(x\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x =4 && a_x=0 && s_x=5 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}5 &= 4t + \frac{1}{2}(0)t^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5=4t\end{align}
Rearranging, we then obtain that \(t=1.25\mbox{ s}\). We now wish to find the value of \(s_y\) at this time and therefore need to use a \(y\) component equation.
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 24.5 && a_y=-9.8 && t=1.25 \end{align}
and we are looking for \(s_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y=u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}s_y&=24.5(1.25)+\frac{1}{2}(-9.8)(1.25^2)\\&\approx 22.97 \mbox{ m}\end{align}
(c) As we determined in part (b), it takes the object \(1.25\) seconds to reach the fence.
Question 12
Two objects, object \(A\) and object \(B\), are projected from \(o\).
They are projected at the same time and with the same speed of \(50\mbox{ m/s}\).
Object \(A\) has an angle of projection of \(60^{\circ}\) and object \(B\) has an angle of projection of \(45^{\circ}\).
Both objects hit the same stationary target. Object \(A\) hits the target at \(t_A\) and object \(B\) hits the target at \(t_B\), where \(t_A \neq t_B\).
(a) Show that \(t_A = \sqrt{2} t_B\).
(b) What is the location of the target relative to \(o\)?
(a) The answer is already in the question!
(b) \(\vec{s} = 186.68\vec{i} + 50.07\vec{j}\mbox{ m}\)
(a)
Object A
\begin{align} u_x = 50\cos 60 && a_x=0 && t=t_A && s=S_x \end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}S_x &= 50 \cos 60^{\circ} t_A + \frac{1}{2}(0)t_A^2\\&=50 \cos 60^{\circ} t_A\end{align}
\[\,\]
Object B
\begin{align} u_x = 50\cos 45 && a_x=0 && t=t_B && s=S_x \end{align}
\begin{align}\downarrow\end{align}
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\begin{align}\downarrow\end{align}
\begin{align}S_x &= 50 \cos 45^{\circ} t_B + \frac{1}{2}(0)t_B^2 \\&= 50 \cos 45^{\circ} t_B\end{align}
\[\,\]
Setting these horizontal displacements equal to each other then gives
\begin{align}50 \cos 60^{\circ} t_A = 50 \cos 45^{\circ} t_B\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t_A & = \frac{\cos 45^{\circ}}{\cos 60^{\circ}}t_B\\& = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}t_B\\&=\sqrt{2} t_B\end{align}
(b)
Object A
\begin{align} u_y = 50\sin 60 && a_y=-9.8 && t=t_A && s=S_y\end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt +\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}S_y = 50 \sin 60^{\circ} t_A+\frac{1}{2}(-9.8)t_A^2\end{align}
\[\,\]
Object B
\begin{align} u_y = 50\sin 45 && a_y=-9.8 && t=t_B && s=S_y \end{align}
\begin{align}\downarrow\end{align}
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\begin{align}\downarrow\end{align}
\begin{align}S_y = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
\[\,\]
Setting these vertical displacements equal to each other then gives
\begin{align}50 \sin 60^{\circ} t_A+\frac{1}{2}(-9.8)t_A^2 = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25 \sqrt{3}t_A -4.9t_A^2 = \frac{50}{\sqrt{2}}t_B – 4.9t_B^2\end{align}
Inserting \(t_A=\sqrt{2}t_B\) results in
\begin{align}25 \sqrt{6} t_B-9.8t_B^2 = \frac{50}{\sqrt{2}}t_B – 4.9t_B^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t_B& \approx \frac{25.88}{4.9}\\&\approx 5.28 \mbox{ s} \end{align}
Inserting it into this
\begin{align}S_x = 50 \cos 45^{\circ} t_B\end{align}
gives
\begin{align}S_x &= 50 \cos 45^{\circ} (5.28)\\&\approx 186.68 \mbox{ m}\end{align}
and inserting it into this
\begin{align}S_y = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
gives
\begin{align}S_y &= 50 \sin 45^{\circ} (5.28)+\frac{1}{2}(-9.8)(5.28^2)\\&\approx 50.07 \mbox{ m}\end{align}
Therefore
\begin{align}\vec{s} = 186.68\vec{i} + 50.07\vec{j}\mbox{ m}\end{align}
(a) We should find an equation for both \(t_A\) and \(t_B\) which occurs at the same connected displacement \(s=S_x\). Let us start with object \(A\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 50\cos 60 && a_x=0 \end{align}
and we are looking for \(t =t_A\) and \(s=S_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}S_x &= 50 \cos 60^{\circ} t_A + \frac{1}{2}(0)t_A^2\\&=50 \cos 60^{\circ} t_A\end{align}
Let us repeat the above approach for object \(B\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_x = 50\cos 45 && a_x=0 \end{align}
and we are looking for \(t =t_B\) and \(s=S_x\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely
\[s_x = u_xt + \frac{1}{2}a_xt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}S_x &= 50 \cos 45^{\circ} t_B + \frac{1}{2}(0)t_B^2 \\&= 50 \cos 45^{\circ} t_B\end{align}
Setting these horizontal displacements equal to each other then gives
\begin{align}50 \cos 60^{\circ} t_A = 50 \cos 45^{\circ} t_B\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t_A & = \frac{\cos 45^{\circ}}{\cos 60^{\circ}}t_B\\& = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}t_B\\&=\sqrt{2} t_B\end{align}
(b) To find \(\vec{s} = s_x\vec{i}+s_y\vec{j}\), we first need to find either \(t_A\) or \(t_B\) so that we can use our result from part (a).
To do this, let us now use \(y\) component equations, starting with object \(A\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 50\sin 60 && a_y=-9.8 \end{align}
and we are looking for \(t =t_A\) and \(s=S_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt +\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}S_y = 50 \sin 60^{\circ} t_A+\frac{1}{2}(-9.8)t_A^2\end{align}
Let us repeat the above approach for object \(B\).
Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!
We are given the following
\begin{align} u_y = 50\sin 45 && a_y=-9.8 \end{align}
and we are looking for \(t =t_B\) and \(s=S_y\).
\[\,\]
Step 2: Choose the most suitable equation(s) based on Step 1.
As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely
\[s_y = u_yt+\frac{1}{2}a_yt^2\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}S_y = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
Setting these vertical displacements equal to each other then gives
\begin{align}50 \sin 60^{\circ} t_A+\frac{1}{2}(-9.8)t_A^2 = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25 \sqrt{3}t_A -4.9t_A^2 = \frac{50}{\sqrt{2}}t_B – 4.9t_B^2\end{align}
If we now insert that \(t_A=\sqrt{2}t_B\), this reduces to the following equation for \(t_B\)
\begin{align}25 \sqrt{6} t_B-9.8t_B^2 = \frac{50}{\sqrt{2}}t_B – 4.9t_B^2\end{align}
Cancelling a \(t_B\) from every term and rearranging gives
\begin{align}t_B& \approx \frac{25.88}{4.9}\\&\approx 5.28 \mbox{ s} \end{align}
We can now insert this into the following equation
\begin{align}S_x = 50 \cos 45^{\circ} t_B\end{align}
to give
\begin{align}S_x &= 50 \cos 45^{\circ} (5.28)\\&\approx 186.68 \mbox{ m}\end{align}
Likewise, we can also insert this into the following equation
\begin{align}S_y = 50 \sin 45^{\circ} t_B+\frac{1}{2}(-9.8)t_B^2\end{align}
to give
\begin{align}S_y &= 50 \sin 45^{\circ} (5.28)+\frac{1}{2}(-9.8)(5.28^2)\\&\approx 50.07 \mbox{ m}\end{align}
Therefore, the displacement fo the target relative to \(o\) is
\begin{align}\vec{s} = 186.68\vec{i} + 50.07\vec{j}\mbox{ m}\end{align}
