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 4 - Kinematics in Two Dimensions - Exercises and Problems: 11

Answer

(a) At t = 0: $v = (2.0\hat{i}+4.0\hat{j})~m/s$ At t = 2 s: $v = (2.0\hat{i})~m/s$ At t = 3 s: $v = (2.0\hat{i}-2.0\hat{j})~m/s$ (b) The magnitude of g is $2.0~m/s^2$ (c) $\theta = 63.4^{\circ}$

Work Step by Step

(a) At t = 1 s: $v = (2.0\hat{i}+2.0\hat{j})~m/s$ At t = 2 s: $v = (2.0\hat{i})~m/s$ At t = 3 s: $v = (2.0\hat{i}-2.0\hat{j})~m/s$ We can see that the vertical component of velocity decreases at a rate of $2.0~m/s^2$. Therefore, at t = 0, $v = (2.0\hat{i}+4.0\hat{j})~m/s$ (b) $v_y = v_{y0}+gt$ $g = \frac{v_y-v_{y0}}{t}$ $g = \frac{2.0~m/s-4.0~m/s}{1.0~s}$ $g = -2.0~m/s^2$ The magnitude of g is $2.0~m/s^2$. (c) We can find the launch angle above the horizontal; $tan(\theta) = \frac{v_y}{v_x}$ $tan(\theta) = \frac{4.0~m/s}{2.0~m/s}$ $\theta = arctan(\frac{4.0~m/s}{2.0~m/s})$ $\theta = 63.4^{\circ}$
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