一只鸟停在距离地面 20 m 的垂直电线杆的顶端 $T$ 处，从地面上的一点 $O$ 观测在 $T$ 处的鸟，仰角为 45°。这只鸟用了一秒钟时间，以恒定的速度水平飞离 $O$，从 $T$ 飞到点 $B$，在整个飞行过程中保持距离地面 20 m 的恒定高度。从 $O$ 观测在 $B$ 处的鸟，其仰角降为 30°。求这只鸟的速度，以 m/s 为单位。 答案 ... m/s

1. Find the horizontal distance from $O$ to the foot of the pole below $T$.

Let the foot of the pole be $P$.

Since $TP=20$ m and the angle of elevation from $O$ to $T$ is $45^\circ$,
$\tan 45^\circ=\frac{20}{OP}$
$OP=\frac{20}{\tan45^\circ}=20$
So the bird starts horizontally $20$ m from $O$.

2. Find the horizontal distance from $O$ to the foot vertically below $B$.

Let the foot vertically below $B$ be $Q$.

The height of $B$ above the ground is still $20$ m, and the angle of elevation from $O$ to $B$ is $30^\circ$.
$\tan30^\circ=\frac{20}{OQ}$
$OQ=\frac{20}{\tan30^\circ}=34.641\ldots$

3. Find the distance flown and hence the speed.

Since the bird flies horizontally away from $O$,
$TB=OQ-OP$
$=34.641\ldots-20$
$=14.641\ldots\text{ m}$

Time taken $=1$ s, so
$\text{speed}=\frac{14.641\ldots}{1}=14.641\ldots\text{ m/s}$
$\text{speed}=14.6\text{ m/s} \quad \text{(3 s.f.)}$