Question

A tennis ball is dropped from a height of 20m it rebounds to a height of 5m

Answer 1

this is the answer to the question

Answer 2

Answer:

The magnitude of average acceleration is 1960 m/s².

Explanation:

Given that,

Height = 20 m

Rebound height = 5 m

Suppose, if the ball is in contact with the floor for 5 millisecond, find the magnitude of average acceleration during the contact

We need to calculate the initial velocity of the ball which is dropped from a height

Using equation of motion

$v^2=u^2+2gh$

$v=\sqrt{2gh}$

Put the value into the formula

$v=\sqrt{2\times9.8\times20}$

$v=19.7\ m/s$

We need to calculate the initial velocity of the ball when it bounced back from the wall and reaches the maximum height of 5 m

Using equation of motion

$v=\sqrt{2gh}$

Put the value into the formula

$v=\sqrt{2\times9.8\times5}$

$v=9.9\ m/s$

We need to calculate the average acceleration

Using formula of average acceleration

$a=\dfrac{v-u}{t}$

Put the value into the formula

$a=\dfrac{9.9-19.7}{5\times10^{-3}}$

$a=-1960\ m/s^2$

Negative sign shows the deceleration.

Hence, The magnitude of average acceleration is 1960 m/s².