Question

A tennis ball is dropped from 1.1 m above the ground. It rebounds to a height of 0.847 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s2 . (Let down be negative.) Answer in units of m/s.

Answer 1

Answer:

Explanation:

it is a inelastic collision

Answer 2

The ball hit the ground with a velocity of 2.2m/s

Given,

The height from which the tennis ball is dropped above the ground = 1.1m

The height rebounded = 0.847m

The acceleration of gravity is 9.8 m/s2

To find,

The ball hit the ground with which velocity

Solution,

Consider,

The height from which the tennis ball is dropped above the ground = 1.1m, to be y1

and

The height rebounded = 0.847m to be y2

The ball hit the ground with which velocity to be Vf

The initial velocity to be Vi

Vi will be equal to zero because it starts from rest.

Vi=0

Use the kinematics equation, which is

$Vf^{2} = Vi^{2} + 2ay$

$Vf^{2}$ = 0 + 2 * 9.8 * (1.1 - 0.847)

Given that the acceleration of gravity is 9.8 m/s2

$Vf^{2}$ = 2 * 9.8 * 0.25

$Vf^{2}$ = 4.9

Vf = √4.9

Vf = 2.2 m/s

The ball hit the ground with a velocity of 2.2m/s

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