质点 $P$ 从水平面上的点 $O$ 以 $24\text{ m s}^{-1}$ 的速率、与水平向上成 $\theta^\circ$ 的方向抛出，随后仅在重力作用下自由运动。在距 $O$ 的水平距离为 $35\text{ m}$ 处，有一堵高 $10\text{ m}$ 的竖直墙，墙面垂直于 $P$ 运动所在的竖直平面。

(a) 求使 $P$ 恰好越过墙的两个 $\theta$ 值。

(b) 已知 $P$ 越过墙，求 $P$ 的落地点距 $O$ 的最小距离。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

把墙顶的坐标代入抛物线轨迹，借助 $\sec^2\theta=1+\tan^2\theta$ 得到关于 $\tan\theta$ 的二次方程。

With $x$ horizontal and $y$ vertically upwards from $O$, the trajectory is

At the top of the wall, $(x,y)=(35,10)$. Writing $z=\tan\theta$ gives

Using the quadratic formula gives

The roots are $z\approx0.769376824$ or $2.522051748$, giving the two angles are $37.6^\circ$ and $68.4^\circ$.

过墙要求在 $x=35$ 的高度至少为 $10$，所以发射角介于上一问两根之间。射程关于发射角先增后减，因此比较两个端点。

The height at the wall is a downward-opening quadratic in $z=\tan\theta$. Therefore the allowed angles lie between the two angles from part (a), including the limiting cases. Setting $y=0$ in the trajectory equation, the positive landing distance is

The function $z/(1+z^2)$ increases up to $z=1$ and decreases thereafter, since its derivative is $(1-z^2)/(1+z^2)^2$. The minimum on the allowed interval is at an endpoint. Using the unrounded roots gives approximately $55.6756\,\mathrm{m}$ and $39.4716\,\mathrm{m}$, so the minimum is $39.5\,\mathrm{m}$.

也可先算飞行时间再乘水平速度。比较两端发射角时，用 $\sin(2\theta)$ 在区间内的增减性来判定最小值。

If $T$ is the flight time, the vertical displacement at landing is zero, so

Taking the nonzero flight time gives $T=4.8\sin\theta$. The horizontal distance is $24T\cos\theta=57.6\sin(2\theta)$. It increases up to $\theta=45^\circ$ and decreases after that. At the two limiting angles the flight times are approximately $2.92696\,\mathrm{s}$ and $4.46205\,\mathrm{s}$. Multiplying each by its corresponding horizontal speed gives approximately $55.6756\,\mathrm{m}$ and $39.4716\,\mathrm{m}$. The minimum is $39.5\,\mathrm{m}$.

若直接使用同高度落地的射程公式，只需比较 $2\theta$ 的两个端点。区间内的唯一驻点给出最大值。

The range formula for landing at the launch height gives

The allowed interval for $2\theta$ contains $90^\circ$ and lies within $(0^\circ,180^\circ)$. Since sine has a maximum at $90^\circ$, its minimum over this interval occurs at an endpoint. Evaluating with the unrounded angles gives $R\approx55.7\,\mathrm{m}$ at the lower angle and $R\approx39.5\,\mathrm{m}$ at the upper angle. Thus the minimum is $39.5\,\mathrm{m}$.