12. After falling from rest from a height of 30-m, a 0.50-kg ball rebounds upward,reaching a maximum height of 20-m. If the contact between the ball and theground lasted 20 milliseconds, what average force was exerted on the ball by theground? (Hint: consider the change in velocity that has occurred between theinterval just before and just after hitting the ground).Answer: 1.1 x 10³ N

Name: 12. After falling from rest from a height of 30-m, a 0.50-kg ball reb
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Description: 12. After falling from rest from a height of 30-m, a 0.50-kg ball rebounds upward,reaching a maximum height of 20-m. If the contact between the ball and theground lasted 20 milliseconds, what average force was exerted on the ball by theground? (Hint

Given data The initial height from where the ball falls is given as h1 = 30 m. The mass of the ball…

Answered: 12. After falling from rest from a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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12. After falling from rest from a height of 30-m, a 0.50-kg ball rebounds upward,
reaching a maximum height of 20-m. If the contact between the ball and the
ground lasted 20 milliseconds, what average force was exerted on the ball by the
ground? (Hint: consider the change in velocity that has occurred between the
interval just before and just after hitting the ground).
Answer: 1.1 x 10³ N

Expert Solution

Step 1

Given data

The initial height from where the ball falls is given as h₁ = 30 m.
The mass of the ball is m = 0.5 kg.
The rebound height attains by the ball is h₂ = 20 m.
The time duration is given as $∆ t = 20 ms$
The initial velocity of the ball is given as u = 0.

The final velocity of the ball for the initial case is calculated as,

$v_{1} = \sqrt{2 g h_{1}} v_{1} = \sqrt{2 \times 9.8 m / s^{2} \times (30 m)} v_{1} = 24.248 m / s$

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