The given situation is shown in the following figure.


If time taken by the ball to reach at highest point \(A\) is \(t\), then
\(0=v-g t \Rightarrow t=\frac{v}{g}\)
Total time taken to reach the ball from point of projection to reach at point \(B\) is given as
\(t_1=t+t=2 t \Rightarrow t_1=\frac{2 v}{g}\)
If \(t_2\) be the time taken by the ball to reach from point \(B\) to point \(C\), then
\(\begin{aligned}
h & =v t_2+\frac{1}{2} g t_2^2 \Rightarrow g t_2^2+2 v t_2-2 h=0 \\
t_2 & =\frac{-2 v \pm \sqrt{4 v^2+8 g h}}{2 g} \\
t_2 & =\frac{-v}{g}+\sqrt{\frac{v^2+2 g h}{2}}
\end{aligned}\)
[taking +ve sign because time is positive]
\(\therefore\) Total time, \(T=t_1+t_2=\frac{2 v}{g}-\frac{v}{g}+\sqrt{\frac{v^2+2 g h}{2}}\)
\(=\frac{v}{g}+\frac{v}{g} \sqrt{1+\frac{2 g h}{v^2}}=\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^2}}\right]\)