Grind #1 - Projectile Motion

A particle is projected from \(o\) on the horizontal ground with an initial velocity of \(20\vec{i}+50\vec{j}\mbox{ m/s}\).

**(a)** How long does it take the object to reach its maximum height?

**(b)** What is the object’s maximum height?

**(c)** How long does it take the object to hit the ground?

**(d)** What is the object’s range?

A particle is projected from \(o\) with a speed of \(25\mbox{ m/s}\).

**(e)** If the object lands \(35\) metres from \(o\), what was the angle of projection?

An object is projected from \(o\) with a speed \(u=35\mbox{ m/s}\) at an angle \(A\) relative to the horizontal ground.

**(f)** 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 \(A\)?

**(g)** What angle of projection will result in the maximum height equalling the range of a projected particle?

Answers

**(a) **\(5.10\mbox{ s}\)

**(b)** \(127.55\mbox{ m}\)

**(c)** \(10.20\mbox{ s}\)

**(d)** \(204\mbox{ m}\)

**(e)** \(16.64^{\circ}\)

**(f)** \(33.82^{\circ}\) and \(88.13^{\circ}\)

**(g)** \(75.96^{\circ}\)

Grind #2 - Differential Equations

Find the general solution of the following:

**(a)** \(\displaystyle\frac{dy}{dx} = \sin x\)

**(b)** \(\displaystyle\frac{dy}{dx} = xy^2\)

**(c)** \(\displaystyle\frac{dy}{dx} = e^{x+y}\)

**(d)** Solve the following differential equation

\begin{align}\frac{dy}{dx} = \frac{8x^3}{y}\end{align}

given that \(y=2\) when \(x=0\).

**(e)** Solve the following differential equation

\begin{align}\frac{d^2y}{dx^2} = 5\frac{dy}{dx}\end{align}

given that both \(y=0\) and \(\displaystyle\frac{dy}{dx}=2\) when \(x=0\).

The rate of decrease of homes available to buy in a city is directly proportional to the size of the housing market.

The number of homes available to buy in this city decreases from \(1{,}000\) to \(800\) over a period of three months.

**(f)** How many homes will be available after a further three months?

Answers

**(a) **\(y(x) = -\cos x+A\)

**(b)** \(y(x) = -\displaystyle\frac{2}{x^2+A}\)

**(c)** \(y(x) = -\ln(-e^x+A)\)

**(d)** \(y(x)= \pm \sqrt{4x^4+4}\)

**(e)** \(y(x) = \dfrac{2}{5} e^{5x} -\dfrac{2}{5}\)

**(f)** \(657\)

Grind #3 - Dynamic Programming

Consider the following multi-stage network:

**(a)** Using dynamic programming, find this network’s *maximum *value.

Consider the following multi-stage network:

**(b)** Using dynamic programming, find this network’s *minimum *value.

Stephanie grows \(4\) kilograms of apples in her garden each week.

Rather than directly selling them as apples, she uses all of them to instead produce three products (juice, sauce and pie) which she then sells.

The amount of profit (in euros) that she makes per kilogram of apples used for each product is as shown.

Stephanie always uses at least \(1\) kilogram of apples to make apple juice each week.

**(c)** Use dynamic programming to determine how Stephanie should use her supply of apples to maximise her profit.

Answers

**(a)Â **\(28\)

**(b)** \(20\)

Stephanie’s maximum profit of â‚¬\(95\) can be achieved in four different ways:

- by allocating \(1\) kilogram of apples to juice, \(1\) kilogram of apples to sauce and the remaining \(2\) kilograms of apples to pie.
- by allocating \(1\) kilogram of apples to juice, \(2\) kilograms of apples to sauce and the remaining \(1\) kilogram of apples to pie.
- by allocating \(2\) kilograms of apples to juice and \(2\) kilograms of apples to pie.
- by allocating \(2\) kilograms of apples to juice, \(1\) kilogram of apples to sauce and the remaining \(1\) kilogram of apples to pie.

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