Answer to Question #134405 in Mechanics | Relativity for John Doe

Question

A tennis ball is dropped from 1.59 m above the ground. It rebounds to a height of 0.921 m.

With what velocity does it hit the ground?

2

021 (part 2 of 3) 10.0 points

With what velocity does it leave the ground? Answer in units of m/s.

022 (part 3 of 3) 10.0 points

If the tennis ball were in contact with the ground for 0.0147 s, find the acceleration given to the tennis ball by the ground.

acceleration of gravity is 9.8 m/s . Answer in units of m.

019 10.0 points

Answer in units of m/s .

The acceleration of gravity is 9.8 m/s . (Let down be negative.)

Answer in units of m/s.

Answer 1

Applying $v^2-u^2=2gH$

$v^2=2\times9.8\times1.59=5.58m/s$

Velocity before hitting the ground is 5.58 m/s

Again applying $v^2-u^2=2gH$ for calculating the velocity with which it leaves the ground

$0-u^2=2\times(-9.8)\times0.921\\u=4.25 m/s$

Velocity with which it leaves the ground is 4.25m/s

Acceleration given to the tennis ball by the ground =$=\frac{v-u}{\Delta t}=\frac{4.25-(-5.58)}{0.0147}=668m/s^2$