Question

a tennis ball is dropped on a on the floor from a height of 20m it rebounds to the height of 5 metre the ball was in contact with the floor for 0.01 second what was its average acceleration during the contact​

Answer 1

Solution:

When it is dropped from a height of 20 m,

$\implies$ Initial height = 20 m

$\implies$ Initial velocity = 0

Velocity (before hitting ground):

$\implies$ √2gh

$\implies$ √2*9.8*20

$\implies$ 19.79 m/s (downward)

Now:

After rebound (maximum height) = 5 m

Final velocity at top = 0

Initial velocity (after rebound):

$\implies$ √2gh

$\implies$ √2*9.8*5

$\implies$ 9.89 m/s (upward)

Assuming downward as positive direction:

Velocity (before hitting ground) = +19.79 m/s 

Velocity (after hitting ground) = -9.89 m/s 

Change in the velocity:

$\implies$ +19.79 - (-9.89)

$\implies$ +19.79 + 9.89

$\implies$ 29.68 m/s

Note: Time = 0.01 s

We know that,

$\implies$ $\boxed{\sf{Acceleration = \frac{Change\:in\:velocity}{Time}}}$

By substituting the values, we get:

$\implies$ 29.68/0.01

$\implies$ 2968 m/s²

_________________

Answered by: Niki Swar, Goa❤️

Answer 2

We know that,

$\implies$ $\boxed{\sf{Acceleration = \frac{Change\:in\:velocity}{Time}}}$

By substituting the values, we get:

$\implies$ 29.68/0.01

$\implies$ 2968 m/s²