Features of Projectile Motion
There are particular features regarding projectile motion that appear frequently in exam questions and are therefore worth highlighting in depth.
We shall discuss each of these features in the most general sense, i.e. without numbers. (At Ordinary Level, it is unlikely that projectile questions will be asked without numbers. We will therefore also consider the features below instead with numbers in the next batch of exercise set questions.)
Therefore, let us consider the general example of a projectile with an initial speed \(u\) and an angle of projection, i.e. direction of the initial velocity vector, \(A\).
Figure 3.3.1
Unless otherwise stated, we shall assume that the object is always projected “northeast”, as in the above diagram.
As such, \(\vec{u_x}\) and \(\vec{u_y}\) will be in same direction as \(\vec{i}\) and \(\vec{j}\) respectively and the corresponding coefficients \(u_x\) and \(u_y\) will be positive.
Therefore, rather than using our usual steps to ensure that we don’t make mistakes with signs, we can instead immediately write down the coefficients as follows:
\begin{align}u_x = u \cos A && u_y = u \sin A\end{align}
Figure 3.3.2
Also, rather than writing \(a_y=-9.8\mbox{ m/s}^2\), we shall instead more generally write it as \(a_y=-g\).
1) What is the projectile’s time of flight?
The time of flight of a projectile is the time it spends in the air from being projected until it hits the ground.
Is there anything special about the SUVAT variables when the projectile hits the ground?
Well, let’s have a look at the displacement vector again.
Figure 3.3.3
We can see that, when the object hits the ground, the displacement in the \(y\) direction is zero!
Key Point
If we are asked to find a projectile’s time of flight, we are being asked to find \(t\) when \(s_y=0\).
Guided Example 3.3.1
An object is projected with a speed \(u\) at an angle \(A\) relative to the horizontal ground.
Show that the time of flight of the object is
\begin{align}t = \frac{2u \sin A}{g}\end{align}
Walkthrough
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 SUVAT 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 \(t=0\) result is expected as \(s_y\) is also zero when the object is initially on the ground at the moment when the object is projected! However, we are instead interested in when it next on the ground, i.e. the time of flight, which is
\begin{align}t=\frac{2 u \sin A}{g}\end{align}
as required.
2) What is the projectile’s range?
The range of a projectile is defined as how far the projectile travels horizontally until it hits the ground.
Again, when the object hits the ground, the displacement in the \(y\) direction is zero. This time, however, we are instead looking for the value of \(s_x\) when this occurs.
Key Point
If we are asked to find a projectile’s range, we are being asked to find \(s_x\) when \(s_y=0\).
Exam Tip
In general, if the specific feature requires finding an \(x\) component based on the condition of a \(y\) component (or vice versa), we should start by first finding the time in which this condition is satisfied using a \(y\) component equation (or vice versa) (as this is the only variable that relates \(x\) components to \(y\) components).
For example, a projectile’s range requires finding an \(x\) component (\(s_x\)) based on the condition of a \(y\) component (\(s_y=0\)). This therefore suggests that we should first find the time in which this \(s_y=0\) condition is satisfied using a \(y\) component equation.
Guided Example 3.3.2
An object is projected with a speed \(u\) at an angle \(A\) relative to the horizontal ground.
Show that the range \(R\) of the object is
\begin{align}R = \frac{u^2 \sin 2 A}{g}\end{align}
Walkthrough
We wish to find the horizontal displacement coefficient \(s_x = R\) when \(s_y=0\).
As stated above, we should first find the time at which \(s_y=0\). Of course, we already found that in the previous example! The time of flight is
\begin{align}t=\frac{2u \sin A}{g}\end{align}
(Note that we cannot simply state this relation when finding the range in an exam question. We need to derive it first as was done in the previous example.)
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 SUVAT 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}
However, notice that this isn’t quite the equation shown in the question. That equation instead has a \(\sin 2 A\) term whereas we have a \(\sin A\cos A\) term. We should therefore have a quick look in our Formula and Tables booklet to see if there is a trigonometric identity that includes both.
Upon inspection, we find the following identity
\begin{align}\sin 2A = 2 \sin A \cos A\end{align}
Inserting this identity therefore results in
\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}
as required.
(As we shall see in the following exercise set, it is often not necessary to perform this last step, and it is sometimes even beneficial not to do it. However, as we were specifically asked to write the equation in this form, we needed to do it here. Writing the range in this form will also be beneficial for the next feature that we will look at.)
Key Point
When being asked to find the range of a projectile, we often need to use the trigonometric identity
\begin{align}\sin 2A= 2 \sin A\cos A\end{align}
3) What angle of projection results in the largest range?
The only thing we can vary in projectile motion is the initial velocity vector – the resulting trajectory from there is all decided by gravity!
We can therefore ask, what initial velocity vector will maximise how far the projectile will go, i.e. how can we maximise the range?
Of course, the magnitude of the initial velocity vector can technically be any value.
However, the direction of the initial velocity vector, i.e. the angle of projection, only has a finite scope. Specifically, if the object is projected “northeast”, it can only take on values above \(0^{\circ}\) and less than \(90^{\circ}\). Which angle within these limits will result in the largest range?
\[\,\]
In the below animation, five objects are projected with the same speed but with different angles of projection (\(15^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\) and \(75^{\circ}\)).
Figure 3.3.4
In this animation, the projectile with an angle of projection of \(45^{\circ}\), i.e. the orange projectile, had the largest range of the five.
Is that the angle we are looking for in general? Well, let’s find out!
Guided Example 3.3.3
An object is projected with a speed \(u\) at an angle \(A\) relative to the horizontal ground.
Show that the range of the object is maximised when \(A=45^{\circ}\).
Walkthrough
To start, we need to first derive the equation for the range. Rather than repeating that process here, we shall skip to stating it. (However, in an exam question, it must first be derived using the previous examples.)
\begin{align}R = \frac{u^2 \sin 2 A}{g}\end{align}
We are therefore asking, what value of \(A\) will make \(R\) in this relation as large as possible?
Well, the only place that \(A\) appears in the above equation is in this term
\begin{align}\sin 2 A\end{align}
Therefore, the question further reduces to asking: what value of \(A\) makes \(\sin 2A\) as large as possible?
Many of you may be aware that the sine function oscillates between values of \(1\) and \(-1\).
Therefore, the maximum possible value of any sine function, including \(\sin 2 A\), is \(1\).
Hence, we therefore wish to find the value of \(A\) such that
\begin{align}\sin 2 A=1\end{align}
Inverting this, we then obtain
\begin{align}2 A&= \sin^{-1} (1)\\&=90^{\circ}\end{align}
and therefore \(A=45^{\circ}\), as required.
In addition, notice that the range of the blue and black projectiles in Figure 3.3.5 are the same, which have angles of projection of \(15^{\circ}\) and \(75^{\circ}\) respectively. This is because \(\sin 2A\) is the same for both angles – check and see!
Likewise, the same is true of the red and green projectiles which have angles of projection of \(30^{\circ}\) and \(60^{\circ}\) respectively.
Finally, note that the time of flight of each of the five projectiles increases as the angle of projection increases. (The blue projectile has the smallest angle of projection and also lands first, the red projectile has the second smallest angle of projection and lands second etc.)
This, again, is as expected. The time of flight formula we derived in Guided Example 3.3.1 is proportional to \(\sin A\), and \(\sin A\) gets larger as \(A\) gets larger for angles between \(0^{\circ}\) and \(90^{\circ}\)!
4) What is the projectile’s maximum height?
This feature is instead completely separate and has nothing to do with when the projectile strikes the ground. Instead, we are being asked about the projectile’s maximum height, i.e. a specific value of \(s_y\).
Is there anything special about the SUVAT variables when the projectile is at its maximum height?
Well, this, time let’s have a look at the velocity vector again.
Figure 3.3.5
Hopefully you can can see that, when the object is at its maximum height, the velocity in the \(y\) direction is zero!
Key Point
If we are asked to find a projectile’s maximum height, we are being asked to find \(s_y\) when \(v_y=0\).
Guided Example 3.3.4
An object is projected with a speed \(u\) at an angle \(A\) relative to the horizontal ground.
Show that the maximum height reached is
\begin{align}H = \frac{u^2 \sin^2 A}{2g}\end{align}
Walkthrough
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 SUVAT 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}
as required.
Of course, we could also ask how long it takes for the object to reach the maximum height. We expect this time to be exactly half the time of flight – the object takes just along travelling up to the maximum height as it does falling back down to the ground. We therefore expect this time to be
\begin{align}t=\frac{u\sin A}{g}\end{align}
However, rather than assume it, let’s prove it!
Guided Example 3.3.5
An object is projected with a speed \(u\) at an angle \(A\) relative to the horizontal ground.
Show that the time taken to reach the maximum height is
\begin{align}t = \frac{u \sin A}{g}\end{align}
Walkthrough
We are looking for the time \(t\) 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 \(t\).
\[\,\]
Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.
As we are not told \(s_y\) nor do we want to find \(s_y\), we should choose the \(y\) component SUVAT equation which does not have \(s_y\) in it, namely
\[v_y=u_y+a_yt\]
\[\,\]
Step 3: Insert the quantities into the equation(s).
Subbing in our known values, we obtain
\begin{align}0=u \sin A -gt\end{align}
Rearranging, this gives
\begin{align}t = \frac{u\sin A}{g}\end{align}
as required.
HIGHER LEVEL ONLY
5) What angle of projection results in striking a target?
We can also be asked questions in which we wish to find the angle of projection \(A\) that results in a projectile of a given initial speed striking a target that is located at some displacement \(\vec{s} = s_x\vec{i}+s_y\vec{j}\).
Note that there is typically two angles of projection that will result in a projectile striking such a target as trajectories can intersect at the location of the target!
Figure 3.3.6
Again, as we are looking for a quantity that involves both \(x\) and \(y\) components, this suggests that we need to find the time \(t\) it takes for the projectile to reach this displacement.
Indeed, that is exactly what our approach should be.
- First, we find the time \(t\) taken to reach this specific value of \(s_x\)
- Then, we find the same time \(t\) taken to reach this specific value of \(s_y\)
- We then set these two times equal to each other and solve for the angle \(A\)
Thankfully, the general case for striking a target is not on the syllabus as the algebra is a bit too messy!
Instead, we shall consider an example involving numbers.
Guided Example 3.3.6
An object is projected from \(o\) with a speed \(u=10\mbox{ m/s}\) at an angle \(A\) relative to the horizontal ground.
If the projectile is to hit a target that is located at \(4\vec{i}+3\vec{j}\mbox{ m}\) from \(o\), what are the two possible values of \(A\)?
Walkthrough
As stated, we first wish to find \(t\) when \(s_x=4\). We shall therefore begin by using 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 = 10\cos A && a_x=0 && s_x=4 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable SUVAT 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}4&=10\cos A t + \frac{1}{2}(0)t^2\\&=10 \cos A t\end{align}
Rearranging, this gives
\begin{align}t = \frac{4}{10\cos A}\end{align}
\[\,\]
Next, we should find \(t\) when \(s_y=3\). 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\sin A && a_y=-9.8 && s_y=3 \end{align}
and we are looking for \(t\).
\[\,\]
Step 2: Choose the most suitable SUVAT 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=10\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 obtain
\begin{align}3&=10\sin A\left(\frac{4}{10\cos A}\right) -\frac{1}{2}(9.8)\left(\frac{4}{10 \cos A}\right)^2\\&=4 \tan A-\frac{0.784}{\cos^2 A} \end{align}
where we have used the identity
\begin{align}\tan A=\frac{\sin A}{\cos A}\end{align}
So, how do we proceed from here? We have the following equation for \(A\)
\begin{align}3 = 4 \tan A- \frac{0.784}{\cos^2 A}\end{align}
but how do we solve it?
Well, ideally, we’d like to replace the \(\cos^2 A\) term with something involving \(\tan A\) so that our equation will then only have \(\tan A\) terms.
This therefore suggests that we should use a trigonometric identity!
A quick look through our Formulae and Tables booklet shows that the most suitable equation is
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
Inserting this identity, we then obtain
\begin{align}3 = 4 \tan A- 0.784(1+\tan^2 A)\end{align}
How does that help? Well, we can now temporarily set \(x=\tan A\) and turn this equation into an algebraic equation that we are more used to!
\begin{align}3=4x-0.784(1+x^2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.784 x^2-4x+3.784=0\end{align}
This can be solved using the quadratic formula to obtain
\begin{align}x&=\frac{4 \pm \sqrt{(-4)^2-4(0.784)(3.784)}}{2(0.784)}\\ &\approx 2.55 \pm 1.30\end{align}
and therefore \(x=3.85\) or \(x=1.25\).
Recall, however, that we’re instead trying to find \(A\) which is related to \(x\) via \(x=\tan A\). Inverting this, we therefore obtain for the first value of \(x\)
\begin{align}A&= \tan^{-1}(3.85)\\&\approx 75.44^{\circ}\end{align}
and for the second value of \(x\)
\begin{align}A&= \tan^{-1}(1.25)\\&\approx 51.34^{\circ}\end{align}
Therefore, if the object is to strike the target, the angle of projection must be either \(75.44^{\circ}\) or \(51.34^{\circ}\).
Key Point
If we are asked to find the angle of projection necessary to strike a target , we often need to use the following trigonometric identities:
\begin{align}\tan A=\frac{\sin A}{\cos A}\end{align}
\begin{align}\frac{1}{\cos^2 A} = 1+\tan^2 A\end{align}
