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Height and Distance
Height and Distance
Medium

Question

A bird is sitting on the top of a vertical pole 2020 m high and its elevation from a point OO on the ground is 45∘{45^ \circ }. It files off horizontally straight away from the point OO. After one second, the elevation of the bird from OO is reduced to 30∘{30^ \circ }. Then the speed (in m/s) of the bird is :

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Solution

Let the speed be yy m/secm/sec. Let ACAC be the vertical pole of height 2020 m.m. Let OO be the point on the ground such that ∠AOC=45∘\angle AOC = {45^ \circ } Let OC=xOC = x Time t=1t=1 ss From ΔAOC,  tan⁡45∘=20x       .....(i)\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right) and from ΔBOD,  tan⁡30∘=20x+y...(ii)\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right) From (i)(i) and (ii),(ii), we have x=20x=20 and 13=20x+y{1 \over {\sqrt 3 }} = {{20} \over {x + y}} ⇒13=2020+y \Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {20 + y}} ⇒20+y=203\Rightarrow 20 + y = 20\sqrt 3 So, y=20(3−1)  i.e.,y = 20\left( {\sqrt 3 - 1} \right)\,\,i.e., speed =20(3−1)m/s = 20\left( {\sqrt 3 - 1} \right)m/s

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