一个质点从水平地面竖直向上抛出。抛出 $2$ 秒后，质点的速率为 $5\,\mathrm{m\,s^{-1}}$，且正向下运动。

(a) 求质点抛出时的速率。

(b) 求质点在两次速率为 $10\,\mathrm{m\,s^{-1}}$ 的时刻之间经过的路程。

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 $10\,\mathrm{m\,s^{-2}}$。

给定的是向下速率5，所以向上取正时速度为负5。

Taking upwards as positive and denoting the projection speed by $u$,

Hence

速率10出现于同一高度的上升、下降两次。由该高度到最高点的距离为5米，总路程为它的两倍。

Let $h$ be the rise from the level where the upward speed is $10\,\mathrm{m\,s^{-1}}$ to the highest point. Then

The particle rises through $h$ and then falls through the same distance before its speed is again $10\,\mathrm{m\,s^{-1}}$. Thus the required distance is

也可用能量求速率10处到最高点的高度差，再利用上、下对称。

If the particle has mass $m$, the kinetic energy at speed $10\,\mathrm{m\,s^{-1}}$ becomes gravitational potential energy at the highest point:

Thus $h=5\,\mathrm{m}$. The two instants with this speed lie on opposite sides of the highest point, at the same height. Therefore the total distance is