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A particle is projected with velocity $2 \sqrt{g h}$ and at an angle $60 °$ to the horizontal so that it just clears two walls of equal height “h” which are at a distance 2h from each other. The time taken by the particle to travel between these two walls is

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

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

$\sqrt{\frac{h}{g}}$

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

answer is D.

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Detailed Solution

$x = u_{x} t = u \cos θ t$

$∴ 2 h = 2 \sqrt{g h} (\cos 60 °) t$

$∴ h = \sqrt{g h} \times \frac{1}{2} t$

$∴ t = 2 \sqrt{\frac{h}{g}}$

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