Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems: 39

Answer

The component of the plane's velocity that is perpendicular to the ground is 15 m/s

Work Step by Step

We can find the slope of the hill; $tan(\theta) = \frac{3}{100}$ $\theta = arctan(\frac{3}{100})$ $\theta = 1.72^{\circ}$ We then find the component of the plane's velocity that is perpendicular to the ground: $v_{perp} = v~sin(\theta)$ $v_{perp} = (500~m/s)~sin(1.72^{\circ})$ $v_{perp} = 15~m/s$ The component of the plane's velocity that is perpendicular to the ground is 15 m/s.
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