一个半径为 $a$、球心为 $O$ 的固定光滑球放在水平地面上。一个质点从球的最高点 $A$ 以速率 $u$ 水平投出。质点开始沿球面在竖直圆周上运动。质点在点 $B$ 脱离球面，其中角 $AOB$ 为 $\theta$。

离开球面后，质点仅在重力作用下运动，随后以速率 $3u$、与水平方向成角 $\beta$ 的方向撞击水平地面（见图）。

求 $\beta$ 的值。

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.

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。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。

从球顶到地面用能量守恒求 $u^2$。离球点的支持力为零，用径向方程与能量守恒求 $\theta$。离开后水平速度保持不变，利用着地速度的水平分量求 $\beta$。

The drop from $A$ to the ground is $2a$, so conservation of energy gives $(3u)^2=u^2+4ga$, hence $u^2=ga/2$.

Let $V$ be the speed at $B$. At loss of contact the normal reaction is zero, so the inward equation is $mg\cos\theta=mV^2/a$. Energy from $A$ to $B$ also gives

Substituting $u^2=ga/2$ and $V^2=ga\cos\theta$ gives

Hence $\cos\theta=5/6$ and $V^2=5ga/6$.

At $B$ the tangent is at angle $\theta$ below the horizontal. During free flight the horizontal velocity is unchanged, so $3u\cos\beta=V\cos\theta$. Therefore

Since the particle strikes the ground moving downward and to the right, $0<\beta<90^\circ$. Thus $\boxed{\beta=69.0^\circ}$ to one decimal place.

离球速度求出后，也可分别算着地时的水平和竖直分量。竖直方向用位移与速度关系，最后用正切求角。

From the energy and radial equations, $V^2=5ga/6$ and $\cos\theta=5/6$. The squared horizontal component stays equal to

The vertical drop from $B$ is $a(1+\cos\theta)$. Taking downward as positive, the vertical component at the ground satisfies

Thus $\tan\beta=\sqrt{847/125}$, giving $\beta=69.0^\circ$ to one decimal place.