College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 64: 71

Answer

$a). 5s$ $b).$ Velocity just before hitting the ground $36.5m/s$

Work Step by Step

u=12.5m/s, S=60m. $v^{2}=u^{2}+2aS$ $0=12.5^{2}-2\times 9.8\times S$ $S=7.97m$ From, $v=u+at$ $0=12.5-9.8t$ $t=1.275s$ Now, total height$=7.97+60=67.97m$ $67.97=0+0.5\times 9.8\times t^{2}$ $t=3.724s$ a). So, total time taken $=1.275s+3.724s=5s$ b). Velocity just before hitting the ground$=u+at=0+9.8\times 3.724=36.5m/s$
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