Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.5 - Vectors: A Geometric Approach - 2.5 Problem Set - Page 107: 37


Magnitude of the velocity $|\textbf{V}|=38.1\,ft/s$ Angle of elevation $\theta=23.2^{\circ}$

Work Step by Step

$|\textbf{V}|=\sqrt {|\textbf{V}_{x}|^{2}+|\textbf{V}_{y}|^{2}}=\sqrt {(35.0\,ft/s)^{2}+(15.0\,ft/s)^{2}}$ $=38.1\,ft/s$ Let the angle of elevation be $\theta$. Then, $\tan\theta=\frac{|\textbf{V}_{y}|}{|\textbf{V}_{x}|}=\frac{15.0}{35.0}$ $\theta=\tan^{-1}(\frac{15.0}{35.0})=23.2^{\circ}$
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