Question

Ball is dropped onto the floor from a height of 20m. it rebounds to a height of 15m. if the ball was in contact with the floor for 0.01sec what was its average acceleration during the contact take g=10m/s

Answer 1

The ball falls to the ground from a height of 20 m. It stays in contact with ground for 0.01 sec. It rebounds to a height of 15 m. We have to find the average acceleration.



Average Acceleration is defined as the change in velocity per unit time.



We have the time of contact. So we simply need to calculate the change in velocity.



The ball initially falls from a height of 20 m. So, just before it hit, the velocity would be:

$v^2 = u^2 + 2as \\ \\ \\ \implies v_{1}^{2} = 0 + 2(10)(20) \\ \\ \\ \implies v_1 = 20 \, \, m/s$


Thus, initial velocity is 20 m/s, in downward direction. 

Ball rebounds. Now just after collision, it acquires some velocity $v_2$ in upward direction. 

We can find it, as we know maximum height is 15 m.

$v^2 = u^2 + 2as \\ \\ \\ \implies 0 = v_{2}^{2} - 2(10)(15) \\ \\ \\ \implies v_{2}^{2} = 300 \\ \\ \\ \implies v_2 = 10\sqrt{3} \, \, m/s \\ \\ \\ \implies v_2 \approx 17.32 \, \, m/s$

Thus, just after collision, the velocity is approximately 17.32 m/s in upward direction. 

If we consider the upward direction as positive, then we have:

$v_1 = -20 \, \, m/s \\ \\ v_2 = +17.32 \, \, m/s$


[Velocity is a vector quantity. So we need to consider direction as well.]


Change in velocity is:

$\Delta v = v_2 - v_1 \\ \\ \\ \implies \Delta v = 17.32 - (-20) \\ \\ \\ \implies \Delta v = 37.32 \, \, m/s$


Now, average acceleration is given by:

$< a > = \frac{\Delta v}{\Delta t} \\ \\ \\ \implies < a > = \frac{37.32}{0.01} \\ \\ \\ \implies \boxed{< a > = 3732 \, \, m / s^2}$


Thus, average acceleration during the contact is approximately 3732 $m / s^2$



Hope it helps
Purva
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