题目

Problem

The points $A$ and $B$ are at the same vertical height $h\,\mathrm{m}$ above horizontal ground. A particle $P$ is released from rest from $A$ . One second later, a particle $Q$ is projected vertically downwards from $B$ with speed $18\,\mathrm{m\,s^{-1}}$ .

Given that $P$ and $Q$ reach the ground at the same time, find the value of $h$ .

(6)

题目中文翻译

点 $A$ 和 $B$ 距水平地面的竖直高度相同，均为 $h\,\mathrm{m}$ 。质点 $P$ 从 $A$ 处由静止释放。一秒后，质点 $Q$ 从 $B$ 处以 $18\,\mathrm{m\,s^{-1}}$ 的速率竖直向下发射。

已知 $P$ 和 $Q$ 同时到达地面，求 $h$ 的值。

(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

解法一

思路

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两质点到地面的位移同为h，但P比Q多运动1秒。统一用Q的运动时间t列方程。

答题过程

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Let $Q$ ‘s time of flight be $t$ seconds. Then $P$ ‘s time of flight is $t+1$ seconds. Taking downwards as positive,

\begin{align*} h=&\,\frac12(10)(t+1)^2,\\[4mm] h=&\,18t+\frac12(10)t^2. \end{align*}

Equating and expanding,

5(t^2+2t+1)=18t+5t^2,

10t+5=18t, \qquad t=\frac58.

Therefore

\begin{align*} h=&\,5\biggl(\frac{13}{8}\biggr)^2=\frac{845}{64}\\[4mm] =&\,13.203125\approx\boxed{13.2\text{ m}}\text{ (3 s.f.)}. \end{align*}

解法二

思路

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也可从Q发射时刻看相对运动：P已领先5m、速度10，之后相同重力加速度使相对速度恒为8。

答题过程

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During $P$ ‘s one-second head start it falls

s_0=\frac12(10)(1)^2=5\text{ m}

and reaches speed $10\,\mathrm{m\,s^{-1}}$ downwards. When $Q$ is projected at $18\,\mathrm{m\,s^{-1}}$ , its closing speed relative to $P$ is $18-10=8\,\mathrm{m\,s^{-1}}$ .

Both have downward acceleration $g$ , so their relative acceleration is zero. They reach the same ground level when $Q$ closes the $5\,\mathrm{m}$ lead:

t=\frac58\text{ s}.

The height is therefore

\begin{align*} h=&\,18\biggl(\frac58\biggr)+5\biggl(\frac58\biggr)^2\\[4mm] =&\,\frac{845}{64}\approx\boxed{13.2\text{ m}}\text{ (3 s.f.)}. \end{align*}

CIE 9709 2026 June Paper 42 Q5

题目

解答

解法一

思路

答题过程

解法二

思路

答题过程