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Question

(C) Let $h$ be the height of the tower. The body is projected upwards with velocity $u$.The time taken to reach the highest point is $t = \frac{u}{g}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $h$ be the height of the tower. The body is projected upwards with velocity $u$.The time taken to reach the highest point is $t = \frac{u}{g}$.The total time taken to reach the ground is $T = n t = \frac{n u}{g}$.Using the equation of motion for displacement $s = ut + \frac{1}{2} a t^2$,where $s = -h$,$u = u$,$a = -g$,and $t = T$:$-h = u T - \frac{1}{2} g T^2$Substitute $T = \frac{n u}{g}$:$-h = u \left( \frac{n u}{g} \right) - \frac{1}{2} g \left( \frac{n u}{g} \right)^2$$-h = \frac{n u^2}{g} - \frac{n^2 u^2}{2 g}$$-h = \frac{2 n u^2 - n^2 u^2}{2 g}$$-h = - \frac{n u^2 (n - 2)}{2 g}$Therefore,the height of the tower is $h = \frac{n u^2 (n - 2)}{2 g}$.

$A$ 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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