在 $t=0$ 时，一个小球从地面上方 $24.5\,\mathrm{m}$ 的 $A$ 点竖直向上抛出。小球首次在 $B$ 点瞬时静止，其中 $AB=19.6\,\mathrm{m}$，并在 $t=T$ 秒时首次撞击地面。

将小球视为仅在重力作用下运动的质点。

(a) 求 $T$ 的值。

(b) 画出小球从 $t=0$ 到 $t=T$ 秒的运动过程的速率—时间图像草图。

（画此草图无需进一步计算。）

先用上升高度求初速，再用从 $A$ 到地面的负位移求总时间。二次方程的负根不对应抛出后的运动，必须排除。

Take upwards as positive and let the initial speed be $u$. At the highest point $B$ the velocity is zero, so

The displacement from $A$ to the ground is $-24.5$ m. Hence

Rejecting $T=-1$ because $T>0$, we obtain $\boxed{T=5}$.

也可将运动分成上升到 $B$ 和从 $B$ 自由下落到地面。最高点离地高度为两段高度之和，两个阶段的时间相加就是总时间。

Let $\tau$ be the time from $A$ to $B$. At the highest point the speed is zero. Using the average velocity during the upward motion,

Therefore $19.6=g\tau^2/2$, giving $\tau=2$ s and $u=19.6\ \mathrm{m\,s^{-1}}$.

The height of $B$ above the ground is $19.6+24.5=44.1$ m. If the fall from rest at $B$ takes $\sigma$ seconds,

Thus $\boxed{T=2+3=5}$.

速率永远非负。上升时速率线性减小至零，下降时线性增大；两段斜率的大小都为 $g$，且下降段持续更久、终点更高。

Draw time $t$ horizontally and speed vertically. The graph is V-shaped:

• It starts above the origin on the speed axis and decreases linearly to zero at the highest point of the motion.
• It then increases linearly to $t=T=5$, ending at a speed greater than the initial speed.
• The two straight segments make equal acute angles with the time axis, since their gradients have magnitudes $g$. The second segment is longer than the first, and the vertex lies on the time axis.

Label the final time $T$ or $5$. The graph stays on or above the time axis.