A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is:

Let us denote by $$O$$ the point on the ground from which the elevations are observed. A vertical pole of height $$20\text{ m}$$ stands with its foot at a point directly below the bird’s initial position.

Initially the bird is perched at the top of this pole. The angle of elevation of the bird from $$O$$ is given as $$45^{\circ}$$. Using the definition of the tangent in a right-angled triangle, we write

$$\tan 45^{\circ}= \frac{\text{opposite side}}{\text{adjacent side}} =\frac{20}{x_0},$$

where $$x_0$$ is the horizontal distance from $$O$$ to the foot of the pole. Since $$\tan 45^{\circ}=1,$$ we obtain

$$1=\frac{20}{x_0}\quad\Longrightarrow\quad x_0=20\text{ m}.$$

Thus, the bird starts $$20\text{ m}$$ horizontally away from the point $$O$$.

The bird now flies off horizontally straight away from $$O$$. Let its constant speed be $$v\text{ m s}^{-1}$$. Because it travels horizontally for exactly $$1\text{ s},$$ the additional horizontal distance covered is

$$v\times 1 = v\text{ m}.$$

Therefore, after one second the total horizontal distance from $$O$$ to the bird becomes

$$x_1 = 20 + v\text{ m}.$$

At this instant the angle of elevation of the bird from $$O$$ is reduced to $$30^{\circ}$$. Applying the tangent definition once more, we have

$$\tan 30^{\circ}= \frac{\text{opposite side}}{\text{adjacent side}} =\frac{20}{20+v}.$$

We know the standard value $$\tan 30^{\circ}= \dfrac{1}{\sqrt{3}}.$$ Substituting this value gives

$$\frac{1}{\sqrt{3}} = \frac{20}{20+v}.$$

Now we cross-multiply to solve for $$v$$ step by step:

$$20+v = 20\sqrt{3}.$$

Isolating $$v$$ yields

$$v = 20\sqrt{3} - 20 = 20(\sqrt{3}-1).$$

Hence the speed of the bird is

$$v = 20(\sqrt{3}-1)\text{ m s}^{-1}.$$

Hence, the correct answer is Option B.