Question

A ball is dropped from a height of 7.2 m. It bounces back to 3.2 m alte
floor. The ball remains in contact with the floor for 20 ms. Given that g=10 ms,
the magnitude of average acceleration of the ball during the contact is
(NSEJS-2011-12)​

Answer 1

The acceleration of the ball is $10^3\ m/s^2$.

Step-by-step explanation:

We have, a ball is dropped from a height of 7.2 m. It bounces back to a height of 3.2 m. Initial and final velocities can be calculated using conservation of energy as :

Initial speed,

$u=\sqrt{2gh}$

h = 7.2 m

$u=\sqrt{2\times 10\times 7.2}=12\ m/s$

Final speed,

$v=\sqrt{2gh'} \\\\v=\sqrt{2\times 10\times 3.2} =8\ m/s$

As it rebounds, v = -8 m/s

The ball remains in contact with the floor for 20 ms, t = 20 ms

It is required to find the average acceleration of the ball during the contact. It is equal to the rate of change of velocity. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{-8-12}{20\times 10^{-3}}\\\\a=-10^3\ m/s^2$

So, the acceleration of the ball is $10^3\ m/s^2$.

Learn more,

Acceleration

https://brainly.in/question/6961927

Answer 2

Step-by-step explanation:

Hope this answer helps you