Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Problems: 56

Answer

3.43 s

Work Step by Step

Let down be the negative direction. First time interval: $x= ut + \frac{1}{2}at^{2}$ $(-9.50)= (0)t + \frac{1}{2}(-9.8)t^{2}$ $t=\sqrt \frac{9.50}{4.9}\approx 1.3924$ s Second time interval: $x=vt-\frac{1}{2}at^{2}$ $(5.70)=(0)t-\frac{1}{2}(-9.8)t^{2}$ $t=\sqrt \frac{5.70}{4.9}\approx 1.0785$ s Third time interval: $x= ut+ \frac{1}{2}at^{2}$ $(-5.70+1.20)= (0)t + \frac{1}{2}(-9.8)t^{2}$ $t=\sqrt \frac{4.50}{4.9}\approx 0.9583$ s Total time = 1.3924+1.0785+0.9583$\; \approx$ 3.43 s
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