Let $u$ be the initial velocity of vertical projection and $t$ be the time taken by the body to reach at height $h$ from the ground.
$\therefore$ Using equation, $h=u t+\frac{1}{2} a t^{2}$
Here, $u=u, a=-g$
$\therefore \quad h=u t-\frac{1}{2} g t^{2}$
$\Rightarrow \quad g t^{2}-2 u t+2 h=0$
$\therefore \quad t=\frac{2 u \pm \sqrt{4 u^{2}-4 g \times 2 h}}{2 g}$
$=\frac{u \pm \sqrt{u^{2}-2 g h}}{g}$
It means $t$ has two values, $t_{1}$ and $t_{2}$. Where, $t_{1}$ is time taken by the body in reaching maximum height and $t_{2}$ be the time taken by the body in reaching ground from maximum height.
$$
\begin{aligned}
&\therefore \quad t_{1}=\frac{u+\sqrt{u^{2}-2 g h}}{g} \\
&\text { and } \quad t_{2}=\frac{u-\sqrt{u^{2}-2 g h}}{g} \\
&\therefore \quad t_{1}+t_{2}=\frac{2 u}{g} \text { or } u=\frac{g\left(t_{1}+t_{2}\right)}{2}
\end{aligned}
$$