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A body projected vertically up with velocity $u$ from the top of a tower reaches the foot of the tower with velocity $2 u .$ Then height of that tower is

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$\frac{u^{2}}{2 g}$

$\frac{u^{2}}{g}$

$\frac{2 u^{2}}{g}$

$\frac{3 u^{2}}{2 g}$

answer is D.

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Detailed Solution

Velocity of the body on reaching foot of the tower is
$V = \sqrt{u^{2} + 2 g h} \Rightarrow 2 u = \sqrt{u^{2} + 2 g h} \Rightarrow 4 u^{2} = u^{2} + 2 g h \Rightarrow h = \frac{3 u^{2}}{2 g}$
So the correct answer is 4

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