Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A ball is thrown vertically downwards from a height of $20 \text{ m}$ with an initial velocity $v_0$ . It collides with the ground, loses $50\%$ of its energy in a collision and rebounds to the same height. The initial velocity $v_0$ is: (Take $g = 10 \text{ m/s}^2$ )

14 m/s

20 m/s

28 m/s

10 m/s

Step-by-Step Solution

Initial Mechanical Energy ( $E_i$ ): At the point of projection (height $h = 20 \text{ m}$ ), the ball possesses both kinetic energy (due to initial velocity $v_0$ ) and potential energy. $E_i = K_i + U_i = \frac{1}{2}mv_0^2 + mgh$
Energy Loss during Collision: The problem states the ball loses $50\%$ of its energy upon impact. Therefore, the energy remaining just after the collision ( $E_f$ ) is $50\%$ of the initial energy. $E_f = 0.5 E_i = 0.5 \left( \frac{1}{2}mv_0^2 + mgh \right)$
Rebound Energy: The ball rebounds to the same height $h = 20 \text{ m}$ . At the maximum height of the rebound, velocity is zero, so the energy is entirely potential. $E_{rebound} = mgh$
Conservation of Energy (Post-Collision): The remaining energy after collision must equal the energy required to reach the rebound height. $0.5 \left( \frac{1}{2}mv_0^2 + mgh \right) = mgh$
Solve for $v_0$ : $\frac{1}{2}mv_0^2 + mgh = 2mgh$ $\frac{1}{2}mv_0^2 = mgh$ $v_0^2 = 2gh$ Substitute $g = 10 \text{ m/s}^2$ and $h = 20 \text{ m}$ : $v_0 = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}$

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