College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 2 - Problems: 21

Answer

(a) $v = 31.0~m/s$ (b) $\theta_x = 58.1^{\circ}$ $\theta_y = 31.9^{\circ}$

Work Step by Step

(a) We can find the magnitude of the velocity vector. $v= \sqrt{v_x^2+v_y^2}$ $v= \sqrt{(16.4~m/s)^2+(-26.3~m/s)^2}$ $v = 31.0~m/s$ (b) We can find the angle $\theta_x$ below the +x-axis. $tan(\theta_x) = \frac{26.3}{16.4}$ $\theta_x = arctan(\frac{26.3}{16.4})$ $\theta_x = 58.1^{\circ}$ We can find the angle $\theta_y$ with the -y-axis. $\theta_y = 90.0^{\circ} - \theta_x$ $\theta_y = 90.0^{\circ} - 58.1^{\circ}$ $\theta_y = 31.9^{\circ}$
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