A body is projected vertically upwards with a velocity ' u ' from the top of a tower. Time taken by it to…

Let us suppose. $h$ is the height of the tower and it attains maximum height up to $H$ from ground, so

Time to reach the maximum height,

$t=\frac{u}{g}$

if $t_2$ is the time taken by the ball to hit the ground,

$-H=u{t}_2-\frac{1}{2}g{\left(t_2\right)}^2$

$t_2=nt$

$$\begin{gathered} -H=u\left(nt\right)-\frac{1}{2}g{\left(nt\right)}^2 \\ \Rightarrow -H=u\left(n\left(\frac{u}{g}\right)\right)-\frac{1}{2}g{\left(n\times \frac{u}{g}\right)}^2 \\ \Rightarrow H=\frac{n{u}^2\left(n-2\right)}{2g} \end{gathered}$$