## Physics (10th Edition)

Published by Wiley

# Chapter 2 - Kinematics in One Dimension - Problems: 52

5.33 m

#### Work Step by Step

$v^{2}=u^{2}+2ax$ $(\frac{u}{2})^{2}=u^{2}+2(-9.8)(4.00)$ $\frac{3}{4}u^{2}=78.4$ $u\approx10.22415$ m/s To find the maximum height reached: $v^{2}=u^{2}+2ax$ $x=\frac{(0)^{2}-(10.22415)^{2}}{2(-9.8)}\approx5.33$ m

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