一个固定实心球的球心为 $O$，半径为 $r$。

质量为 $m$ 的质点 $P$ 被保持在球的光滑外表面上的 $A$ 点，$OA$ 与竖直向上方向成 $\alpha$ 角，其中 $\alpha<45^\circ$。

如图 4 所示，将质点由静止释放后，它在 $B$ 点以速率 $v$ 离开球面，$OB$ 与水平面成 $\alpha$ 角。

假设空气阻力可忽略。

(a) 证明 $v^2=2gr(\cos\alpha-\sin\alpha)$。

(b) 证明 $\tan\alpha=\dfrac23$。

$P$ 经过过 $O$ 的水平线的瞬间，其运动方向与水平面成 $\theta$ 角。

(c) 证明 $\cos\theta=\dfrac2{\sqrt{39}}$。

$A$ 点的角度从竖直量起，$B$ 点的角度从水平量起，因此两点高度分别为 $r\cos\alpha$ 与 $r\sin\alpha$。支持力不做功，用下降的高度差求动能。

Relative to the horizontal through $O$, the heights of $A$ and $B$ are $r\cos\alpha$ and $r\sin\alpha$ respectively. The particle starts from rest and the normal reaction does no work, so

Cancelling $m$ and multiplying by $2$ gives

离开外球面时，向外的支持力降为零，重力的向心分量独自提供向心力。结合(a)消去速率，才可推出角度关系。

At $B$, the particle leaves the sphere, so the normal reaction is zero. Resolving towards $O$ gives

Using part (a),

so

Dividing by $3\cos\alpha$,

离开球面后水平速度保持不变。用从 $A$ 到过 $O$ 水平面的总能量求末速率，再取水平分量与速率之比，避免先把角度换成小数。

Let $C$ be the point where the projectile crosses the horizontal through $O$, and let its speed there be $W$. From part (b),

At $B$, the velocity is tangent to the sphere, downwards and to the right. Its horizontal component is $v\sin\alpha$. This stays constant during the flight, so

By part (a),

Conservation of energy from rest at $A$ to $C$ gives

Squaring (1) and using these exact expressions,

Since $\theta$ is the acute angle below the horizontal,

另一条路线是分开计算落至 $C$ 时的水平、竖直速度。竖直方向可用常加速度公式，因为离开球面后仅受重力。

From part (b), $\sin\alpha=2/\sqrt{13}$ and $\cos\alpha=3/\sqrt{13}$; part (a) gives $v^2=2gr/\sqrt{13}$.

The initial horizontal and downward vertical velocity components at $B$ are $v\sin\alpha$ and $v\cos\alpha$. The horizontal component remains constant, so its square at $C$ is

The fall from $B$ to $C$ is $r\sin\alpha$. Applying the vertical constant-acceleration equation,

Therefore

As $\theta$ is acute,