A ball is dropped on the floor from a height of $10 ext{ m}$. It rebounds to a height of $2.5 ext{ m}$. If

Question

A ball is dropped on the floor from a height of $10 	ext{ m}$. It rebounds to a height of $2.5 	ext{ m}$. If the ball is in contact with the floor for $0.01 	ext{ sec}$, the average acceleration during contact is:

Accepted Answer

1.  **Velocity just before impact ($v_i$):** The ball falls freely from height $h_1 = 10 	ext{ m}$. Using the kinematic equation $v^2 = u^2 + 2gh$ with initial velocity $u=0$ and $g=9.8 	ext{ m/s}^2$:
    $v_i = \sqrt{2gh_1} = \sqrt{2 	imes 9.8 	imes 10} = \sqrt{196} = 14 	ext{ m/s}$.
    Taking the upward direction as positive, $v_i = -14 	ext{ m/s}$ (downwards) .
2.  **Velocity just after rebound ($v_f$):** The ball rises to a height $h_2 = 2.5 	ext{ m}$. Using $v^2 = u^2 - 2gh$ where the final velocity at the peak is 0:
    $0 = v_f^2 - 2(9.8)(2.5) \implies v_f = \sqrt{49} = 7 	ext{ m/s}$.
    Since it is rebounding, $v_f = +7 	ext{ m/s}$ (upwards) .
3.  **Change in Velocity ($\Delta v$):**
    $\Delta v = v_f - v_i = (+7) - (-14) = 21 	ext{ m/s}$.
4.  **Average Acceleration:** Average acceleration is the change in velocity divided by the time interval ($\Delta t = 0.01 	ext{ s}$).
    $a_{avg} = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 	ext{ m/s}^2$ .
5.  **Direction:** Since the change in velocity is positive, the acceleration is directed upwards.

Step-by-Step Solution

Practice All PHYSICS Questions