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Q.

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

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a

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

b

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

c

$\frac{\sqrt{2 H sin θ}}{g}$

d

$\frac{2 \sqrt{2 H sin θ}}{g}$

answer is B.

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Detailed Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

and $T = \frac{2 u \sin θ}{g}$

$∴$ $\frac{T^{2}}{H} = {(\frac{2 u \sin θ}{g})}^{2} \times \frac{2 g}{u^{2} \sin^{2} θ} = \frac{8}{g}$

or $T = 2 \sqrt{\frac{2 H}{g}}$ .

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