The maximum height reached by a projectile is $h$ . It's time of flight is

\sqrt{\frac{4 h}{g}}

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\frac{8 h}{g}

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\sqrt{\frac{8 h}{g}}

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\sqrt{\frac{16 h}{g}}

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Solution

The correct option is C $\sqrt{\frac{8 h}{g}}$
Maximum height, $H = \frac{u^{2} {sin}^{2} θ}{2 g}$
$\Rightarrow u sin θ = \sqrt{2 g h}$
Time of flight, $T = \frac{2 u sin θ}{g}$
$= \frac{2}{g} \cdot \sqrt{2 g h}$
$= \sqrt{\frac{8 h}{g}}$

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