Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Problems - Page 50: 50


12 m

Work Step by Step

$v^{2}=u^{2}+2ax$ $u^{2}=v^{2}-2ax$ $u^{2}=(0)^{2}-2(-9.8)(16)=313.6$ Half of the initial speed = $\frac{\sqrt{ 313.6}}{2}$ $v^{2}=u^{2}+2ax$ $x=\frac{v^{2}-u^{2}}{2a}$ $x=\frac{(\frac{\sqrt 313.6}{2})^{2}-(313.6)}{2(-9.8)}=12$ m
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