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Question: A particle is projected with velocity \(2\sqrt{gh}\) and at an angle \({{60}^{\circ }}\) with the ho...

A particle is projected with velocity 2gh2\sqrt{gh} and at an angle 60∘{{60}^{\circ }} with 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
A. 22hg\text{A}\text{. }2\sqrt{\dfrac{2h}{g}}
B. h2g\text{B}\text{. }\sqrt{\dfrac{h}{2g}}
C. 2hg\text{C}\text{. }2\sqrt{\dfrac{h}{g}}
D. hg\text{D}\text{. }\sqrt{\dfrac{h}{g}}

Explanation

Solution

Resolve the initial velocity of the particle into its vertical and horizontal components. The horizontal component will remain constant throughout the particle’s motion since there is no force acting on the particle in that direction. With this knowledge find the time taken for the particle to travel between the walls.

Formula used:
time = distancespeed\text{time = }\dfrac{\text{distance}}{\text{speed}}

Complete answer:
It is given that a particle is projected with a velocity of 2gh2\sqrt{gh} making an angle of 60∘{{60}^{\circ }} with the horizontal.
Now, resolve the velocity of the particle in the horizontal and vertical components. The horizontal component will be 2ghcos⁡60∘=2gh(12)=gh2\sqrt{gh}\cos {{60}^{\circ }}=2\sqrt{gh}\left( \dfrac{1}{2} \right)=\sqrt{gh}
And the vertical component will be 2ghsin⁡60∘=2gh(32)=3gh2\sqrt{gh}\sin {{60}^{\circ }}=2\sqrt{gh}\left( \dfrac{\sqrt{3}}{2} \right)=\sqrt{3gh}.
The particle will be accelerated downwards due to the force of gravity. Hence, the vertical component of the velocity will change with time.
Since there is no force acting on the particle in the horizontal direction, the horizontal component of the velocity will not change. Therefore, it will remain constant through the motion of the particle.
It is given that the distance between the two walls is 2h. Let the time taken to cross both the walls be t. This means that during the time t, the horizontal displacement of the particle is 2h. And the speed with which it covers this distance is gh\sqrt{gh}.
Let use the formula time = distancespeed\text{time = }\dfrac{\text{distance}}{\text{speed}}.
⇒t=2hgh=2hg\Rightarrow t=\dfrac{2h}{\sqrt{gh}}=2\sqrt{\dfrac{h}{g}}

Hence, the correct option is C.

Note:
There are many ways to solve the given problem. This was the shortest solution and easiest solution to the problem.
Another way is to find the times when the particle is at the tips of both the walls and then find the difference between these times. The difference will be the time taken by the particle to travel between the walls.