Question

A particle is projected with velocity $2\sqrt{gh}$ and at an angle ${60}^{\circ }$ the horizontal so that it just clear two walls of equal height ${}^{\prime }{h}^{\prime }$ which are at a distance $2h$ from each other. The time taken by the particle to travels between these two walls is

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

Answer 1

Answer: (C)

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

Given - $u=2\sqrt{gh},\theta ={60}^o,$ horizontal distance $d=2h$

As the particle is projected with an initial velocity , at an angle with horizontal ,therefore it is a projectile motion . We know that , the horizontal velocity remains constant in a projectile motion .

Hence , horizontal velocity of particle is

$u_x=u\cos \theta$

or $u_x=2\sqrt{gh}\times \cos 60$

or $u_x=2\sqrt{gh}\times 1/2=\sqrt{gh}$

Therefore , time taken by particle to cover the horizontal distance of $2h$ will be ,

$t=2h/u_x=2h/\sqrt{gh}=2\sqrt{h/g}$