Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 203: 14c


2.98 m/s

Work Step by Step

As in Problem 14b, we use: $v = \sqrt {v_{0}^{2} – 2g\Delta h}$ Here the change in height will be $\Delta h$ = 0, because the ball is at exactly the same height it was at initially. The initial speed will be the one found in Problem 14a, $v_{0}$ = 2.98 m/s. So plugging in our values, we get: $v = \sqrt {v_{0}^{2} – 2g\Delta h}$ = $\sqrt {(2.98 m/s)^{2} – 2(9.8 m/s²)(0)}$ = 2.98 m/s
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