Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems: 6

Answer

(a) The minimum initial speed 12.9 m/s (b) The speed when the ball hits the ground is 14 m/s

Work Step by Step

(a) Let $KE_1$ be the initial kinetic energy. Let $PE_1$ be the potential energy at height $h_1=1.5~m$. Let $KE_2$ be the kinetic energy at maximum height. Let $PE_2$ be the potential energy at maximum height $h_2 = 10~m$. We can use conservation of energy to find the minimum speed $v_1$. $KE_1+PE_1 = KE_2+PE_2$ $\frac{1}{2}mv_1^2+mg~h_1 = 0+mg~h_2$ $v_1^2 = 2g~(h_2-h_1)$ $v_1 = \sqrt{2g~(h_2-h_1)}$ $v_1 = \sqrt{(2)(9.80~m/s^2)(10~m-1.5~m)}$ $v_1 = 12.9~m/s$ The minimum initial speed 12.9 m/s (b) Let $KE_3$ be the kinetic energy when the ball hits the ground. Let $PE_3$ be the potential energy at $h_3 = 0$. We can use conservation of energy to find the speed $v_3$ when the ball hits the ground. $KE_3+PE_3 = KE_2+PE_2$ $\frac{1}{2}mv_3^2+0 = 0+mg~h_2$ $v_3^2 = 2g~h_2$ $v_3 = \sqrt{2g~h_2}$ $v_3 = \sqrt{(2)(9.80~m/s^2)(10~m)}$ $v_3 = 14~m/s$ The speed when the ball hits the ground is 14 m/s
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