Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 54: 90

Answer

(a) $2.8~s$ (b) $-29.5~m/s$

Work Step by Step

(a) Let the ground be the origin. $x_0=45~m$, $x=0$, $v_0=-2.0~m/s$, $a=-g=-9.81~m/s^2$ $x=x_0+v_0t+\frac{1}{2}at^2$ $0=45~m+(-2.0~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$ $(4.905~m/s^2)t^2+(2.0~m/s)t-45~m=0$ This is a quadratic equation for $t$. $t=\frac{-(2.0~m/s)±\sqrt {(2.0~m/s)^2-4(4.905~m/s^2)(-45~m)}}{2(4.905~m/s^2)}$ $t_1=2.8~s$ $t_2=-3.2~s$ (this solution is not valid) (b) $v=v_0+at$ $v=-2.0~m/s+(-9.81~m/s^2)(2.8~s)=-29.5~m/s$
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