JEE Main · 2014 · 06 Apr 2014 · Offline · Heights and Distances

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

4 5^{\circ}

. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to

3 0^{\circ}

. Then the speed (in m/s) of the bird is

(A) $40 (2 - 1)$
(B) $40 (3 - 2)$
(C) $202$
(D) $20 (3 - 1)$ ✓ Correct

Answer: (D)

Solution

At first the bird is

20

m up and seen at

4 5^{\circ}

, so the horizontal distance to it is

20 cot 4 5^{\circ} = 20

m. A second later it is still

20

m up but seen at

3 0^{\circ}

, so the new horizontal distance is

20 cot 3 0^{\circ} = 203

m. It has moved

203 - 20 = 20 (3 - 1)

m in one second, so its speed is

20 (3 - 1)

m/s.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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