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: 76


$a). v=2gt,h=gt^{2}$ $b).$ total maximum height$=3gt^{2}$ $c).$ Maximum height $=26460m$

Work Step by Step

a) a=2g, $v=0+at=2gt$ $h=0+\frac{1}{2}2g\times t^{2}=gt^{2}$ b). $v^{2}=u^{2}+2aS$ $0=(2gt)^{2}-2gS$ $S=2gt^{2}$ So, total maximum height$=2gt^{2}+gt^{2}=3gt^{2}$ c). t=30s, Maximum height $=3\times 9.8 \times 900=26460m$
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