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A small body is projected horizontally from the top of a tower of height h with velocity 30 m/s. When the body is at a height h/2 above the ground its velocity vector makes an angle of $45^{o}$ with horizontal. Then h is equal to (Take $g = 10 m / s^{2}$ )

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120 m

60 m

45 m

90 m

answer is D.

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Detailed Solution

At a height h/2 above the ground horizontal component of velocity $V_{1} = 30 m / s$

Vertical component of velocity $V_{2} = \sqrt{2 . g . h / 2} = \sqrt{gh}$

$Now, \frac{V_{2}}{V_{1}} = \tan 45^{o} \Rightarrow \frac{\sqrt{10 . h}}{30} = 1 \Rightarrow h = 90 m$

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