Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Summary Exercises on Applications of Trigonometry and Vectors: 4

Answer

The balloon's speed will be 15.8 ft/s The balloon's path will make an angle of $71.6^{\circ}$ with the horizontal.

Work Step by Step

The hot-air balloon's resultant vector is $(5.00~ft/s, 15.0~ft/s)$ We can find the magnitude of the vector which is the balloon's speed: $v = \sqrt{(5.00~ft/s)^2+(15.0~ft/s)^2}$ $v = 15.8~ft/s$ The balloon's speed will be 15.8 ft/s We can find the angle above the horizontal: $tan~\theta = \frac{15.0}{5.00}$ $\theta = arctan(\frac{15.0}{5.00})$ $\theta = 71.6^{\circ}$ The balloon's path will make an angle of $71.6^{\circ}$ with the horizontal.
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