Question

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

• A. $2 \sqrt{\frac{2 h}{g}}$
• B. $\sqrt{\frac{h}{2 g}}$
• C. $2 \sqrt{\frac{h}{g}}$
• D. $\sqrt{\frac{h}{g}}$

Answer 1

Answer: (C)

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, $t=2 \sqrt{\frac{h}{g}}$