question_answer

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 \[45{}^\circ\]. 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:   JEE Main  Solved  Paper-2014

A) \[40\left( \sqrt{2}-1 \right)\]                      

B) \[40\left( \sqrt{3}-\sqrt{2} \right)\]

C)  \[20\sqrt{2}\]                  

D) \[20\left( \sqrt{3}-1 \right)\]

Correct Answer: D

Solution :

Here, \[AP=QB=20m\] \[\angle POA={{45}^{o}},\angle QOB={{30}^{o}}\] \[\Rightarrow \]\[OA=20;OB=20\sqrt{3}\] \[\Rightarrow \]\[OB-OA=20(\sqrt{3}-1)\] Hence distance covered in one second by the bird is \[AB=20(\sqrt{3}-1)\] Thus, speed of bird \[=20(\sqrt{3}-1)m/s\]