A stone is thrown at an angle $θ$ to the horizontal reaches a maximum height H. Then the time of flight of stone will be

\sqrt{\frac{H}{g}}

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2 \sqrt{\frac{2 H}{g}}

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\frac{2 \sqrt{2 H s i n θ}}{g}

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\frac{\sqrt{2 H s i n θ}}{g}

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Solution

The correct option is B $2 \sqrt{\frac{2 H}{g}}$
By using $s = v t - \frac{1}{2} a t^{2},$
We get $H = 0 - \frac{1}{2} (- g) {(\frac{T}{2})}^{2}$

$\Rightarrow T = 2 \sqrt{\frac{2 H}{g}}$

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