题目

Problem

A particle $P$ is projected with speed $u$ at an angle $\alpha$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of $P$ from $O$ at a subsequent time $t$ are denoted by $x$ and $y$ respectively.

(a) Derive the equation of the trajectory of $P$ in the form

y=x\tan\alpha-\frac{gx^2}{2u^2}\sec^2\alpha.

(3)

During its flight, $P$ must clear an obstacle of height $h\text{ m}$ that is at a horizontal distance of $32\text{ m}$ from the point of projection. When $u=40\sqrt2\text{ m s}^{-1}$ , $P$ just clears the obstacle. When $u=40\text{ m s}^{-1}$ , $P$ only achieves $80\%$ of the height required to clear the obstacle.

(b) Find the two possible values of $h$ .

(6)

题目中文翻译

一个质点 $P$ 从水平面上的点 $O$ 以速率 $u$ 、与水平方向成仰角 $\alpha$ 抛出，此后仅在重力作用下自由运动。此后时刻 $t$ ， $P$ 相对于 $O$ 的水平位移和竖直位移分别记为 $x$ 和 $y$ 。

(a) 推导 $P$ 的轨迹方程，并写成以下形式：

y=x\tan\alpha-\frac{gx^2}{2u^2}\sec^2\alpha.

(3)

在飞行过程中， $P$ 必须越过一个高度为 $h\text{ m}$ 、与抛出点的水平距离为 $32\text{ m}$ 的障碍物。当 $u=40\sqrt2\text{ m s}^{-1}$ 时， $P$ 恰好越过障碍物。当 $u=40\text{ m s}^{-1}$ 时， $P$ 仅达到越过障碍物所需高度的 $80\%$ 。

(b) 求 $h$ 的两个可能值。

(6)

原卷通用作答要求

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 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

展开

分别写水平和竖直位移。水平运动匀速，可用水平式消去时间。

答题过程

展开

The horizontal and vertical displacement equations are

\begin{align*} x=&\,ut\cos\alpha,\\[4mm] y=&\,ut\sin\alpha-\frac12gt^2. \end{align*}

The first equation gives $t=x/(u\cos\alpha)$ . Substituting into the second gives

\begin{align*} y=&\,u\sin\alpha\frac{x}{u\cos\alpha} -\frac g2\bigg(\frac{x}{u\cos\alpha}\bigg)^2\\[4mm] =&\,x\tan\alpha-\frac{gx^2}{2u^2}\sec^2\alpha, \end{align*}

as required.

(b)

解法一

思路

展开

在相同横坐标处写出两种初速对应的高度。用较小初速的高度等于 $0.8h$ 消去 $h$ ，再分别代入求障碍物高度。

答题过程

展开

Let $z=\tan\alpha$ and use $\sec^2\alpha=1+z^2$ . At $x=32$ with $g=10$ , the two trajectories give

\begin{align*} h=&\,32z-\frac85(1+z^2),\\[4mm] \frac45h=&\,32z-\frac{16}{5}(1+z^2). \end{align*}

Multiplying the first equation by $4/5$ and equating the expressions for $4h/5$ gives

\begin{align*} \frac{128}{5}z-\frac{32}{25}(1+z^2) =&\,32z-\frac{16}{5}(1+z^2),\\[4mm] \frac{32}{5}z=&\,\frac{48}{25}(1+z^2),\\[4mm] 10z=&\,3(1+z^2),\\[4mm] 3z^2-10z+3=&\,0,\\[4mm] (3z-1)(z-3)=&\,0. \end{align*}

Thus $z=1/3$ or $z=3$ . Substituting into the first height equation gives

\begin{align*} z=\frac13:\quad h=&\,\frac{32}{3}-\frac85\frac{10}{9} =\frac{80}{9}\,\mathrm{m},\\[4mm] z=3:\quad h=&\,96-\frac85\times10=80\,\mathrm{m}. \end{align*}

CIE 9231 2023 November Paper 32 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程