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: 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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.