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A bird is sitting on the top of a vertical pole
20m 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 $(m / s)$ of the bird is

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$40 (\sqrt{2} - 1)$

$40 (\sqrt{3} - \sqrt{2})$

$20 \sqrt{2}$

$20 (\sqrt{3} - 1)$

answer is D.

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Detailed Solution

Let $P Q$ be the pole of height 20 m $P O Q = 45^{\circ}$

Let $P'$ be the position of the bird after one second

$P^{'} O Q^{'} = 30^{\circ}$

$\begin{array}{l} P P^{'} = Q Q^{'} = P Q^{'} - P Q = 20 \cot 30^{\circ} - 20 \cot 45^{\circ} \\ = 20 (\sqrt{3} - 1) \end{array}$

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