Q188 / 200 · Manipal (MET) 2010 · Physics

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

A
$\sqrt{\frac{2 H}{g}}$
Correct answer
B
$2 \sqrt{\frac{2 H}{g}}$
C
$\frac{2 \sqrt{2 H \sin \theta}}{g}$
D
$\frac{\sqrt{2 H \sin \theta}}{g}$

Solution

$H=\frac{u^2 \sin ^2 \theta}{2 g}$ and $T=\frac{2 u \sin \theta}{g}$ $\Rightarrow T^2=\frac{4 u^2 \sin ^2 \theta}{g^2}$ $\therefore \frac{T^2}{H}=\frac{8}{g}$ $\Rightarrow T=\sqrt{\frac{8 H}{g}}=2 \sqrt{\frac{2 H}{g}}$

Hence, the correct answer is option: 'B'.

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