一只鸟停在离地面 20 米高的垂直电线杆顶部 $T$ 处，从地面上的一点 $O$ 仰望 $T$ 处鸟的仰角为 $45^\circ$。这只鸟用了一秒钟的时间，以恒定速度水平飞行，从 $T$ 飞到远离 $O$ 的点 $B$，整个飞行过程中保持离地面 20 米的恒定高度。从 $O$ 到 $B$ 处鸟的仰角减小到 $30^\circ$。求这只鸟的飞行速度（以 m/s 为单位）。 答：... m/s

1. Find the horizontal distance from O to the pole (let's call the base of the pole P).

Let $P$ be the point on the ground directly below $T$.
In $\triangle OPT$, $\angle OPT = 90^\circ$, $TP = 20$ m, and $\angle TOP = 45^\circ$.
$\tan 45^\circ = \frac{TP}{OP}$
$OP = \frac{20}{\tan 45^\circ} = 20$ m

2. Find the horizontal distance from O to the point directly below B (let's call it Q).

Let $Q$ be the point on the ground directly below $B$.
In $\triangle OQB$, $\angle OQB = 90^\circ$, $BQ = 20$ m, and $\angle BOQ = 30^\circ$.
$\tan 30^\circ = \frac{BQ}{OQ}$
$OQ = \frac{20}{\tan 30^\circ} = 20\sqrt{3}$ m

3. Calculate the distance flown and the speed.

Distance flown $TB = PQ = OQ - OP = 20\sqrt{3} - 20$ m.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{20\sqrt{3} - 20}{1} = 20(\sqrt{3} - 1) \approx 14.641$ m/s