## Physics (10th Edition)

Published by Wiley

# Chapter 2 - Kinematics in One Dimension - Problems: 51

3.2 m

#### Work Step by Step

Let up be the positive direction. a) $v^{2}=u^{2}+2ax$ $v=±\sqrt {(u^{2}+2ax)}$ $v=±\sqrt {((1.8)^{2}+2(-9.8)(-3.0))}$ $v\approx-7.9$ m/s since final velocity is in the negative direction b) $v^{2}=u^{2}+2ax$ $x=\frac{v^{2}-u^{2}}{2a}$ $x=\frac{(0)^{2}-(1.8)^{2}}{2(-9.8)}\approx0.2$ m Hence, the distance above the water = 3.0 + 0.2 = 3.2 m

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