Projectile motion is the motion of a launched object, subject to only the constant acceleration of gravity, $g\text{ ms}^{-2}$. The object is called a projectile, and its path is called a trajectory. A projectile is launched from the origin with an initial velocity of $u\text{ ms}^{-1}$ at an angle $\theta$ radian above the horizontal, where $0 < \theta < \frac{\pi}{2}$, as shown in the diagram above. The trajectory of a projectile depends on its motion in two dimensions - horizontal ($x$) and vertical ($y$) components. The position of the projectile at time $t$ seconds after it is launched can be defined by
$x = ut \cos \theta \quad \text{and} \quad y = ut \sin \theta - \frac{1}{2}gt^2,$
where $u$ and $g$ are positive real constants.

1. Set the vertical displacement $y$ to $0$ and factorise the equation.

$ut \sin \theta - \frac{1}{2}gt^2 = 0$
$t \left(u \sin \theta - \frac{1}{2}gt\right) = 0$

2. Solve for the non-zero value of $t$.

Since $t \neq 0$,
$u \sin \theta - \frac{1}{2}gt = 0$
$\frac{1}{2}gt = u \sin \theta$
$t = \frac{2u\sin\theta}{g}$

3. Interpret the physical meaning of this value of $t$ in context.

This value of $t$ represents the total time of flight of the projectile (i.e. the time taken for the projectile to return to the same horizontal level from which it was launched).

1. Determine the time of flight $t$ when the projectile returns to its initial height, based on part (a).

Setting $y = 0$:
$ut \sin \theta - \frac{1}{2}gt^2 = 0$
$t \left( u \sin \theta - \frac{1}{2}gt \right) = 0$
For the return to the initial height, $t \ne 0$, so:
$u \sin \theta - \frac{1}{2}gt = 0 \implies t = \frac{2u \sin \theta}{g}$

2. Substitute $t$ into the equation for the horizontal distance $x$.

$x = ut \cos \theta$
$x = u \left( \frac{2u \sin \theta}{g} \right) \cos \theta$
$x = \frac{2u^2 \sin \theta \cos \theta}{g}$

3. Use the double angle formula to complete the proof.

Applying the trigonometric identity $2 \sin \theta \cos \theta = \sin(2\theta)$:
$x = \frac{u^2}{g} \sin(2\theta) \quad \text{(shown)}$

1. Differentiate the range $x$ with respect to $\theta$.

$\frac{\mathrm{d}x}{\mathrm{d}\theta} = \frac{u^2}{g}(2\cos 2\theta) = \frac{2u^2}{g}\cos 2\theta$

2. Equate the derivative to zero to find the stationary point.

$\text{For maximum range, } \frac{\mathrm{d}x}{\mathrm{d}\theta} = 0$
$\frac{2u^2}{g}\cos 2\theta = 0$
$\cos 2\theta = 0$

3. Solve the trigonometric equation for $\theta$.

Since $0 < \theta < \frac{\pi}{2}$, we have $0 < 2\theta < \pi$.
$2\theta = \frac{\pi}{2}$
$\theta = \frac{\pi}{4}$

4. Perform the second derivative test to confirm it is a maximum.

$\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2} = \frac{2u^2}{g}(-2\sin 2\theta) = -\frac{4u^2}{g}\sin 2\theta$
When $\theta = \frac{\pi}{4}$, $\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2} = -\frac{4u^2}{g}\sin\left(\frac{\pi}{2}\right) = -\frac{4u^2}{g}$.
Since $u$ and $g$ are positive, $\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2} < 0$, which confirms that the range is maximum at $\theta = \frac{\pi}{4}$.

5. Find the maximum range by substituting $\theta = \frac{\pi}{4}$ back into the equation for $x$.

$\text{Maximum range } x = \frac{u^2}{g}\sin\left(2 \times \frac{\pi}{4}\right)$
$x = \frac{u^2}{g}\sin\left(\frac{\pi}{2}\right) = \frac{u^2}{g}(1) = \frac{u^2}{g}$