Question

A ball falls in the downward direction from a height \( h \) with initial velocity \( V \). It collides with the ground, loses \( \frac{3}{4} \)th of energy and comes back to the same height. The initial velocity \( V \) is (g = acceleration due to gravity)

• A. \( \sqrt{gh} \)
• B. \( \sqrt{6gh} \)
• C. \( \sqrt{3gh} \)
• D. \( \sqrt{2gh} \)

Hint:

When a ball loses energy during a collision, the remaining energy can be used to calculate the velocity required to reach the same height.

Answer 1

The Correct Option is B

Step 1: Energy loss during collision.
When the ball collides with the ground, it loses \( \frac{3}{4} \) of its energy. Therefore, the remaining energy after the collision is \( \frac{1}{4} \) of its initial potential energy. Step 2: Use the potential energy formula.
The initial potential energy \( U \) at the height \( h \) is: \[ U = Mgh \] where \( M \) is the mass of the ball and \( g \) is the acceleration due to gravity. The velocity required to reach the same height after losing \( \frac{3}{4} \) of energy is found from the remaining energy. Step 3: Apply the kinematic equation.
The velocity \( V \) needed for the ball to reach height \( h \) after the energy loss is given by: \[ V = \sqrt{6gh} \] Step 4: Conclusion.
Thus, the initial velocity \( V \) is \( \sqrt{6gh} \), which corresponds to option (B).