抛体运动是指发射物体的运动，其仅受重力加速度常数 $g\text{ ms}^{-2}$ 的作用。该物体被称为抛体，其路径被称为轨迹。 如上图所示，一个抛体从原点以大小为 $u\text{ ms}^{-1}$ 的初速度发射，其发射角与水平方向的夹角为 $\theta$ 弧度，其中 $0 < \theta < \frac{\pi}{2}$。 抛体的轨迹取决于它在两个维度（水平分量 $x$ 和竖直分量 $y$）上的运动。抛体在发射后 $t$ 秒时的位置可以定义为：
$x = ut \cos \theta \quad \text{and} \quad y = ut \sin \theta - \frac{1}{2}gt^2,$
其中 $u$ 和 $g$ 为正实常数。

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 $x$.

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