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 53: 81

Answer

Bill: $x=3.0~m-(4.905~m/s^2)t^2$ Ted: $x=1.0~m+(4.2~m/s)t-(4.905~m/s^2)t^2$

Work Step by Step

$x=x_0+v_0t+\frac{1}{2}at^2$ Bill: $x_0=3.0~m$, $v_0=0$, $a=-g=-9.81~m/s^2$ $x=3.0~m+0t+\frac{1}{2}(-9.81~m/s^2)t^2$ $x=3.0~m-(4.905~m/s^2)t^2$ Ted: $x_0=1.0~m$, $v_0=4.2~m/s$, $a=-g=-9.81~m/s^2$ $x=1.0~m+(4.2~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$ $x=1.0~m+(4.2~m/s)t-(4.905~m/s^2)t^2$
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