Answer: Option C
:
C

Time of flight for this part of motion (between points A and B, both at height h) assuming speed of projection to be $v$ and angle of projection to be $\alpha$
$T=\frac{2vsin\alpha }{g}$ ---------(I)
Now we have to eliminate $v$ & $\alpha$ as these variables we assumed.
Hmmmmm..........How do we do that, okay what else is given?
Yes, the maximum height i.e. $(H-h)$.
$\Rightarrow$ $(H-h)$ = $\frac{v^{2}si{n}^2\alpha }{2g}$ ----------(II)
Thank goodness!
Now from equation (I) $\Rightarrow$ $vsin\alpha$ = $\frac{Tg}{2}$
Equation (II) $\Rightarrow$ $(H-h)$ = $\frac{T^2{g}^2}{4\left(2g\right)}$
$\Rightarrow$$(H-h)$ = $\frac{g{T}^2}{8}$
$\Rightarrow$ $T$ = $\sqrt{\frac{8(H-h)}{g}}$