Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 64: 20

Answer

a)$\vec v =-24.4 m/s$ b)$t =4.53 s$

Work Step by Step

lets take upward direction as positive. Thus we have $\vec u = 20m/s \\ \vec s = -10m \\ \vec a =-g =-9.8m/s^2 \\ \vec v =? \\ t =?$ a) We know that $2\vec a. \vec s = v^2 - u^2$ $2*(-9.8)*(-10) = v^2 - 20^2$ $v^2 =196 +400$ $ v=±\sqrt 596m/s$ since the velocity will be in downward direction. Therefore $\vec v =-24.4 m/s$ b) We have $\vec v = \vec u + \vec a t $ $ -24.4 =20 -9.8t$ $ t =\frac{ 20+24.4}{9.8}$s $t =4.53 s$
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