Physics (10th Edition)

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

Chapter 3 - Kinematics in Two Dimensions - Problems - Page 74: 24

Answer

3.4 m

Work Step by Step

Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the traveled time of the criminal. $\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation. $-2\space m=0+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$ $t=0.64\space s$ Let's apply the equation $S=ut$ in the horizontal direction to find the horizontal distance traveled after launch. $\rightarrow S=ut$ ; Let's plug known values into this equation. $D=5.3\space m/s\times 0.64\space s=3.4\space m$ Horizontal distance traveled after launch = 3.4 m
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