Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Summary Exercises on Applications of Trigonometry and Vectors - Page 345: 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.
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.