Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 47


The ball will reach a height of 2h

Work Step by Step

The potential energy at the maximum height is equal to the kinetic energy at the initial position. $PE_1 = KE_0$ $mgh = KE_0$ $h = \frac{KE_0}{mg}$ Let's suppose that the kinetic energy is $2KE_0$ at the initial position and let $h_2$ be the height the ball reaches. $PE_2 = 2KE_0$ $mg~h_2 = 2KE_0$ $h_2 = \frac{2KE_0}{mg}$ $h_2 = 2~\frac{KE_0}{mg}$ $h_2 = 2~h$ The ball will thus reach a height of 2h.
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