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(C) Let $t_1$ be the time taken to reach the maximum height from the top of the tower. At the maximum height, the final velocity is $0$. Using the equ

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$\frac{n u^2(n-2)}{2 g}$

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(C) Let $t_1$ be the time taken to reach the maximum height from the top of the tower. At the maximum height, the final velocity is $0$. Using the equation $v = u + at$, we get $0 = u - gt_1$, which gives $t_1 = \frac{u}{g}$.Let $t_2$ be the total time taken to reach the ground. According to the problem, $t_2 = n t_1 = \frac{nu}{g}$.Using the equation of motion for displacement $s = ut + \frac{1}{2}at^2$, where $s = -H$ (downward displacement), $u$ is the initial upward velocity, $a = -g$, and $t = t_2$:$-H = u t_2 - \frac{1}{2} g t_2^2$Substituting $t_2 = \frac{nu}{g}$:$-H = u \left( \frac{nu}{g} \right) - \frac{1}{2} g \left( \frac{nu}{g} \right)^2$$-H = \frac{nu^2}{g} - \frac{n^2 u^2}{2g}$$-H = \frac{2nu^2 - n^2u^2}{2g} = -\frac{nu^2(n-2)}{2g}$Therefore, $H = \frac{nu^2(n-2)}{2g}$.

$A$ body is projected vertically upwards with a velocity $u$ from the top of a tower. The time taken by it to reach the ground is $n$ times the time taken by it to reach the highest point in its path. The height of the tower is:

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