Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 6 - Work and Energy - Problems - Page 169: 58



Work Step by Step

According to the work-energy theorem, $$E_f-E_0=W_{nc}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{nc}$$ We take the initial point to be when the bat hits the ball and the final point to be 25m above the point of impact. So $v_0=40m/s$, $h_f-h_0=25m$, $m=0.14kg$ and $W_{nc}=70J$. Therefore, $$0.07(v_f^2-1600)+1.372(25)=70$$ $$0.07(v_f^2-1600)=35.7$$ $$v_f^2=2110$$ $$v_f=45.93m/s$$
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