一个小球从水平地面上方 $19.6\,\mathrm{m}$ 的 $A$ 点，以速率 $29.4\,\mathrm{m\,s^{-1}}$ 竖直向上抛出。

在撞击地面之前，将小球视为仅在重力作用下运动的质点。假设小球不反弹。

(a) 求小球速率小于 $14.7\,\mathrm{m\,s^{-1}}$ 的过程中所经过的路程。

(b) 求小球速率大于 $29.4\,\mathrm{m\,s^{-1}}$ 的持续时间。

(c) 画出小球从在 $A$ 点被抛出的瞬间到撞击地面的瞬间的速率—时间图像草图，清楚标出图像与坐标轴的交点。

速率小于给定值的过程横跨最高点，包括上升和下降两段。先算从上升速率 $14.7$ 到最高点的距离，再利用运动对称性把这段路程加倍。

Let $s$ be the distance travelled from the point where the upward speed is $14.7\ \mathrm{m\,s^{-1}}$ to the highest point. Using $v^2=u^2+2as$,

After the highest point, the ball falls the same distance before its downward speed reaches $14.7\ \mathrm{m\,s^{-1}}$.

Thus the total distance travelled while its speed is below $14.7\ \mathrm{m\,s^{-1}}$ is

小球返回 $A$ 时向下速率恰好为初速 $29.4$。只有从此之后到落地的这一段速率更大，所以以返回 $A$ 为新的计时起点。

When the ball returns to the level of $A$, its downward speed is $29.4\ \mathrm{m\,s^{-1}}$. Its speed exceeds this value during the subsequent fall of 19.6 m to the ground.

Let this interval last $\tau$ seconds. Taking downwards as positive,

Rejecting the negative root, the required duration is

也可先求落地速率，再求速度从向下 $29.4$ 增大到落地速率所需的时间。全程使用向下为正，避免把初始上抛方向混入最后一段。

On returning to $A$, the ball has downward speed $29.4\ \mathrm{m\,s^{-1}}$. Let its downward speed at the ground be $v_g$. Then

During the interval when the speed exceeds $29.4=3g$, the acceleration is $g$ downwards. Hence

速率图是 V 形，最低点对应最高位置。先求到最高点的时间，标出初始速率与时间轴交点，下降段末端速率高于初速。

The time to reach the highest point is

Draw a speed-time graph starting at $(0,29.4)$ and decreasing linearly to $(3,0)$. It then increases linearly until the ball hits the ground, ending at a speed greater than 29.4.

The two segments have gradients $-g$ and $+g$, so they make equal acute angles with the time axis. Label both axis intercepts clearly. Do not include a vertical drop at the final instant.