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: 68


$a). 2.29 s$ $b). 16.5 m/s$

Work Step by Step

$v^{2}=u^{2}+2aS$ $0=6^{2}-2\times 9.8\times S$ $S=1.84m$ So, the stone goes upto a height of 1.84 m. So, the stone has to travel =$1.84+12=13.84m$ before it hits the ground. Now, $v=u+at$ or, $0=6-9.8t$ i.e. $t=0.61s$ $S=ut+\frac{1}{2}at^{2}$ $13.84=0+0.5\times 9.8t^{2}$ $t=1.68s$ a). So, time of total flight $=0.61s+1.68s=2.29s$ b). Also, velocity just before hitting the ground$=u+at=0+9.8\times1.68=16.5m/s$
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