如图所示，一个物体从平台顶部的 $O$ 点以与水平方向成 $\theta$ 的抛射角抛出，其中 $0 < \theta < \frac{\pi}{2}$，初始速度为 $u\text{ ms}^{-1}$。$x$-$y$ 平面包含该物体的运动轨迹，以通过 $O$ 点的水平轴和竖直轴为参考，该轨迹的方程为 $y = x \tan \theta - \frac{10x^2}{2u^2 \cos^2 \theta},$ 其中 $x\text{ m}$ 和 $y\text{ m}$ 分别是如上图所示从 $O$ 点向右和向上移动的距离。 点 $P$ 在水平地面上，位于 $O$ 点正下方 $1.8\text{ m}$ 处，而 $Q$ 是物体落地的点。

1. Determine the coordinates $(x, y)$ of the landing point $Q$.

Since $P$ is $1.8\text{ m}$ vertically below the origin $O$, and $Q$ is $15\text{ m}$ horizontally from $P$ on the ground, the coordinates of $Q$ are $x = 15$ and $y = -1.8$.

2. Substitute the coordinates of $Q$ and the angle of projection into the trajectory equation.

Substitute $x = 15$, $y = -1.8$, and $\theta = \frac{\pi}{4}$ into the trajectory equation $y = x \tan \theta - \frac{10x^2}{2u^2 \cos^2 \theta}$:
$-1.8 = 15 \tan\left(\frac{\pi}{4}\right) - \frac{10(15)^2}{2u^2 \cos^2\left(\frac{\pi}{4}\right)}$
Using $\tan\left(\frac{\pi}{4}\right) = 1$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ (so $\cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2}$):
$-1.8 = 15(1) - \frac{2250}{2u^2 \left(\frac{1}{2}\right)}$
$-1.8 = 15 - \frac{2250}{u^2}$

3. Solve the equation for the initial speed $u$.

Rearrange the equation to make $u^2$ the subject:
$\frac{2250}{u^2} = 15 + 1.8$
$\frac{2250}{u^2} = 16.8$
$u^2 = \frac{2250}{16.8} = \frac{1875}{14} \approx 133.93$
Since speed $u > 0$, we take the positive square root:
$u = \sqrt{\frac{1875}{14}} \approx 11.5727$
$u = 11.6\text{ m s}^{-1} \text{ (to 3 s.f.)}$

1. Substitute $u = 10$ and $y = -1.8$ into the trajectory equation.

At point $Q$, the object is on the ground, which is $1.8\text{ m}$ below the origin $O$. Thus, the vertical displacement is $y = -1.8$. We are also given $u = 10$. Substituting these into the trajectory equation:
$-1.8 = x \tan \theta - \frac{10x^2}{2(10)^2 \cos^2 \theta}$
$-1.8 = x \left(\frac{\sin \theta}{\cos \theta}\right) - \frac{10x^2}{200 \cos^2 \theta}$
$-1.8 = x \frac{\sin \theta}{\cos \theta} - \frac{x^2}{20 \cos^2 \theta}$

2. Multiply by $20 \cos^2 \theta$ and rearrange the equation to form a quadratic in $x$.

Multiply the entire equation by $20 \cos^2 \theta$ to eliminate the denominators:
$-36 \cos^2 \theta = 20x \sin \theta \cos \theta - x^2$
Rearranging all terms to one side to make the $x^2$ coefficient positive:
$x^2 - 20x \sin \theta \cos \theta - 36 \cos^2 \theta = 0$

3. Apply double-angle identities for sine and cosine to obtain the final equation.

Using the double-angle identities $\sin 2\theta = 2 \sin \theta \cos \theta$ and $\cos 2\theta = 2 \cos^2 \theta - 1$ (which rearranges to $2\cos^2\theta = \cos 2\theta + 1$):
$x^2 - 10(2 \sin \theta \cos \theta) - 18(2 \cos^2 \theta) = 0$
$x^2 - 10x \sin 2\theta - 18(\cos 2\theta + 1) = 0$
$x^2 - 10x \sin 2\theta - 18 \cos 2\theta - 18 = 0 \quad \text{(shown)}$

1. Differentiate the given equation implicitly with respect to $\theta$.

Given the equation from part (ii):
$x^2 - 10x \sin 2\theta - 18 \cos 2\theta - 18 = 0$

We differentiate each term implicitly with respect to $\theta$, taking care to use the product rule for the $-10x \sin 2\theta$ term:
$2x \frac{dx}{d\theta} - \left(10 \frac{dx}{d\theta} \sin 2\theta + 10x(2 \cos 2\theta)\right) - 18(-2 \sin 2\theta) - 0 = 0$
$2x \frac{dx}{d\theta} - 10 \frac{dx}{d\theta} \sin 2\theta - 20x \cos 2\theta + 36 \sin 2\theta = 0$

2. Apply the condition for a stationary value of $x$.

At the stationary value of $x$, the derivative $\frac{dx}{d\theta} = 0$. Substituting this into our differentiated equation:
$2x(0) - 10(0) \sin 2\theta - 20x \cos 2\theta + 36 \sin 2\theta = 0$
$-20x \cos 2\theta + 36 \sin 2\theta = 0$

3. Rearrange the equation to express $x$ in terms of $\tan 2\theta$.

Move the terms to opposite sides to isolate $x$:
$20x \cos 2\theta = 36 \sin 2\theta$
$x = \frac{36 \sin 2\theta}{20 \cos 2\theta}$
$x = \frac{9}{5} \tan 2\theta$
$x = 1.8 \tan 2\theta$

1. Derive the result of part (iii) by implicitly differentiating the equation to find an expression for the stationary value of $x$.

$x^2 - 10x \sin 2\theta - 18 \cos 2\theta - 18 = 0$

Differentiating implicitly with respect to $\theta$:
$2x \frac{dx}{d\theta} - \left( 10 \frac{dx}{d\theta} \sin 2\theta + 20x \cos 2\theta \right) - 18(-2 \sin 2\theta) = 0$
$2x \frac{dx}{d\theta} - 10 \frac{dx}{d\theta} \sin 2\theta - 20x \cos 2\theta + 36 \sin 2\theta = 0$

At the stationary value of $x$, $\frac{dx}{d\theta} = 0$:
$-20x \cos 2\theta + 36 \sin 2\theta = 0 \implies 20x \cos 2\theta = 36 \sin 2\theta$
$x = \frac{36 \sin 2\theta}{20 \cos 2\theta} = 1.8 \tan 2\theta$

2. Substitute $x = 1.8 \tan 2\theta$ into the original equation to form a trigonometric equation in terms of $\theta$.

$(1.8 \tan 2\theta)^2 - 10(1.8 \tan 2\theta) \sin 2\theta - 18 \cos 2\theta - 18 = 0$
$3.24 \frac{\sin^2 2\theta}{\cos^2 2\theta} - 18 \frac{\sin^2 2\theta}{\cos 2\theta} - 18 \cos 2\theta - 18 = 0$

Divide by 18:
$0.18 \frac{\sin^2 2\theta}{\cos^2 2\theta} - \frac{\sin^2 2\theta}{\cos 2\theta} - \cos 2\theta - 1 = 0$

3. Simplify the trigonometric equation into a quadratic in terms of $\cos 2\theta$ and solve for its roots.

Multiply the equation by $\cos^2 2\theta$:
$0.18 \sin^2 2\theta - \sin^2 2\theta \cos 2\theta - \cos^3 2\theta - \cos^2 2\theta = 0$

Substitute $\sin^2 2\theta = 1 - \cos^2 2\theta$:
$0.18(1 - \cos^2 2\theta) - \cos 2\theta(1 - \cos^2 2\theta) - \cos^3 2\theta - \cos^2 2\theta = 0$
$0.18 - 0.18 \cos^2 2\theta - \cos 2\theta + \cos^3 2\theta - \cos^3 2\theta - \cos^2 2\theta = 0$
$-1.18 \cos^2 2\theta - \cos 2\theta + 0.18 = 0$

Multiply by $-50$ to achieve integer coefficients:
$59 \cos^2 2\theta + 50 \cos 2\theta - 9 = 0$

Factorising the quadratic equation:
$(59 \cos 2\theta - 9)(\cos 2\theta + 1) = 0$

Since $0 < \theta < \frac{\pi}{2}$, we have $0 < 2\theta < \pi$, which means $\cos 2\theta \neq -1$. Thus:
$\cos 2\theta = \frac{9}{59}$

4. Calculate the angle $\theta$ and find the corresponding stationary value of $x$.

$2\theta = \cos^{-1}\left(\frac{9}{59}\right) \approx 1.4176 \text{ rad}$
$\theta \approx 0.7088 \approx 0.71 \text{ rad (to 2 d.p.)}$

Substitute $2\theta \approx 1.4176$ back to find $x$:
$x = 1.8 \tan(1.4176) \approx 11.6619$
$x \approx 11.7 \text{ (to 3 s.f.)}$