College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 7

Answer

The maximum height of the swing is $~2.3~meters~$ above the ground.

Work Step by Step

We can use conservation of energy to find the height $h_f$ when the swing reaches its maximum height: $U_f+KE_f= U_0+KE_0$ $mgh_f+0 = mgh_0+\frac{1}{2}mv_0^2$ $h_f = h_0+\frac{v_0^2}{2g}$ $h_f = (0.5~m)+\frac{(6.0~m/s)^2}{(2)(9.80~m/s^2)}$ $h_f = 2.3~m$ The maximum height of the swing is $~2.3~meters~$ above the ground.
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