质点 $P$ 从水平地面竖直向上抛出，达到的最大高度为 $45\,\mathrm{m}$。落地后，$P$ 停止运动且不反弹。

(a) 求 $P$ 抛出时的速率。

(b) 求 $P$ 的速率至少为 $10\,\mathrm{m\,s^{-1}}$ 的总时间。

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 $10\,\mathrm{m\,s^{-2}}$。

最高点速度为零，用速度与位移关系求初速。

Taking upwards as positive, let the projection speed be $u$. At the maximum height,

Since $u>0$, the projection speed is $\boxed{30\,\mathrm{m\,s^{-1}}}$.

速率至少10包括向上速度不小于10，以及向下速度不大于−10，两段都要计算。

While $P$ is airborne, its velocity is $v=30-10t$. It reaches the ground again at $t=6$, since

On the upward journey, $v\ge10$ gives $30-10t\ge10$, hence $0\le t\le2$. On the downward journey, $v\le-10$ gives $30-10t\le-10$, hence $4\le t<6$.

The total time for which the speed is at least $10\,\mathrm{m\,s^{-1}}$ is