Question

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 ${45}^0$. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to ${30}^0$. Then the speed (in m/sec) of the bird is -----

Answer 1

Let the speed of the bird is $xm/s$

From figure

$\tan 30=\frac{20}{20+x}=\frac{1}{\sqrt{3}}$

$20+x=20\sqrt{3}$

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

Therefore, speed of the bird is $20(\sqrt{3}-1)m/s$