Trigonometry (10th Edition)

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

Chapter 7 - Review Exercises: 40

Answer

The magnitude of the horizontal component of $v$ is $25\sqrt{2}$ The magnitude of the vertical component of $v$ is $25\sqrt{2}$

Work Step by Step

We can find the magnitude of the horizontal component of $v$: $v_x = v~cos~\theta$ $v_x = (50)~cos~45^{\circ}$ $v_x = (50)~(\frac{\sqrt{2}}{2})$ $v_x = 25\sqrt{2}$ The magnitude of the horizontal component of $v$ is $25\sqrt{2}$ We can find the magnitude of the vertical component of $v$: $v_y = v~sin~\theta$ $v_y = (50)~sin~45^{\circ}$ $v_y = (50)~(\frac{\sqrt{2}}{2})$ $v_y = 25\sqrt{2}$ The magnitude of the vertical component of $v$ is $25\sqrt{2}$
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