A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 450. It flies off horizontally straight away from the point O. After one second, the elevation of bird from O is reduced to 300, then the speed (in m/s), of the bird is
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answer is 0014.64.
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Detailed Solution
Let AB= height of vertical pole =20 From ΔOAB,tan45∘=ABOA⇒OA=AB=20 Let C be the new position of bird Let BC=x From ΔOCD,tan30∘=20x+20⇒x=20(3−1)=14.64 m/s