A body when projected at an angle ' $\theta$ ' with the horizontal reaches a maximum height ' H '. The time of flight of the body will be ( $g=$ acceleration due to gravity)

Time of flight in terms of maximum height

The maximum height $H$ and time of flight $T$ for a projectile with initial velocity $u$ at angle $\theta$ are given by:

$H = \frac{u^2 \sin^2\theta}{2g}$

$T = \frac{2u \sin\theta}{g}$

From the height equation, $u^2 \sin^2\theta = 2gH$, so $u \sin\theta = \sqrt{2gH}$.

Substituting into the time equation yields:

$T = \frac{2}{g} \cdot \sqrt{2gH} = 2\sqrt{\frac{2H}{g}}$

This corresponds to option C.