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A stone is thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
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Verified Answer
The correct answer is:
$2 \sqrt{\frac{2 H}{g}}$
$H=\frac{u^2 \sin ^2 \theta}{2 g}$ and $T=\frac{2 u \sin \theta}{g} \Rightarrow T^2=\frac{4 u^2 \sin ^2 \theta}{g^2}$
$\therefore \frac{T^2}{H}=\frac{8}{g} \Rightarrow T=\sqrt{\frac{8 H}{g}}=2 \sqrt{\frac{2 H}{g}}$
$\therefore \frac{T^2}{H}=\frac{8}{g} \Rightarrow T=\sqrt{\frac{8 H}{g}}=2 \sqrt{\frac{2 H}{g}}$
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