A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in \(t_1\). If it is…

Case 1: \(h=-u{t}_{1}+\frac{1}{2}g{t}_{1}^{2}\) ( downward taken positive)

Case 2 \(h=u{t}_{2}+\frac{1}{2}g{t}_{2}^{2}\)

\begin{equation}
\begin{aligned}
& \text { } t_1=\frac{u+\sqrt{u^2+2 g h}}{g} \\
& t_2=\frac{-u+\sqrt{u^2+2 g h}}{g} \\
& \text{Case 3}\\ t=\frac{\sqrt{2 g h}}{g} \\
& t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2 \\
& \Rightarrow t=\sqrt{t_1 t_2}
\end{aligned}
\end{equation}