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A particle is projected with velocity $2 \sqrt{g h}$ and at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2 h$ from each other. The time taken by the particle to travel between these two walls is
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Verified Answer
The correct answer is:
$2 \sqrt{\frac{h}{g}}$
Given, velocity of particle $=2 \sqrt{g h}$
Angle $(\theta)=60^{\circ}$
Distance covered by particle $=2 h$
Now,
$$
2 h=2 \sqrt{g h} \cos 60^{\circ} t
$$
or,
$\sqrt{h}=\sqrt{g} \times \frac{1}{2} t$
or,
$t=2 \sqrt{\frac{h}{g}}$
Angle $(\theta)=60^{\circ}$
Distance covered by particle $=2 h$
Now,
$$
2 h=2 \sqrt{g h} \cos 60^{\circ} t
$$
or,
$\sqrt{h}=\sqrt{g} \times \frac{1}{2} t$
or,
$t=2 \sqrt{\frac{h}{g}}$
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