Question

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the average acceleration of

Answer 1

Acceleration,
`767.08\ m/s^2`

Step-by-step explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

`v^2=u^2+2ah`

u = initial velocity=0 m/s

a = acceleration=g

`v=√(2gh) \\\\v=√(2* 9.8* 17.3) \\\\v=18.41\ m/s`

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

`a=(v-u)/(t)\\\\a=(0-18.41)/(24* 10^(-3))\\\\a=767.08\ m/s^2`

So, the average acceleration of the ball during the time it is in contact is
`767.08\ m/s^2`.