A ball is thrown vertically downwards from a height of $20 ext{ m}$ with an initial velocity $v_0$. It colli

Question

A ball is thrown vertically downwards from a height of $20 	ext{ m}$ 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 $g = 10 	ext{ m/s}^2$)

Accepted Answer

1. **Initial Mechanical Energy ($E_i$):** At the point of projection (height $h = 20 	ext{ m}$), the ball possesses both kinetic energy (due to initial velocity $v_0$) and potential energy.
   $$ E_i = K_i + U_i = \frac{1}{2}mv_0^2 + mgh $$
2. **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.
   $$ E_f = 0.5 E_i = 0.5 \left( \frac{1}{2}mv_0^2 + mgh ight) $$
3. **Rebound Energy:** The ball rebounds to the *same height* $h = 20 	ext{ m}$. At the maximum height of the rebound, velocity is zero, so the energy is entirely potential.
   $$ E_{rebound} = mgh $$
4. **Conservation of Energy (Post-Collision):** The remaining energy after collision must equal the energy required to reach the rebound height.
   $$ 0.5 \left( \frac{1}{2}mv_0^2 + mgh ight) = mgh $$
5. **Solve for $v_0$:**
   $$ \frac{1}{2}mv_0^2 + mgh = 2mgh $$
   $$ \frac{1}{2}mv_0^2 = mgh $$
   $$ v_0^2 = 2gh $$
   Substitute $g = 10 	ext{ m/s}^2$ and $h = 20 	ext{ m}$:
   $$ v_0 = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20 	ext{ m/s} $$

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