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: 38

Answer

The arrow was shot at 29.2 ft/s at an angle of elevation of $59.0^{\circ}$.

Work Step by Step

$|\textbf{V}|=\sqrt {|\textbf{V}_{x}|^{2}+|\textbf{V}_{y}|^{2}}=\sqrt {(15.0\,ft/s)^{2}+(25.0\,ft/s)^{2}}$ $=29.2\,ft/s$ Let the angle of elevation be $\theta$. Then $\tan\theta=\frac{|\textbf{V}_{y}|}{|\textbf{V}_{x}|}=\frac{25.0}{15.0}$ $\implies \theta=\tan^{-1}(\frac{25.0}{15.0})=59.0^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.