Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 85: 28b

Answer

$|\vec{v} |=27.6$ m/s

Work Step by Step

(b) The horizontal component of velocity remains unchanged, so at A, $v_{x}=v_{0x}=v_{0}\cos\theta_{0}=(42.0m/s)\cos 60^{o}=21.3m/s$ The vertcal component is given by Eq.4-23 $v_{y}=v_{0y}\sin\theta_{0}-gt=(42.0m/s)\sin 60^{o}-(9.80m/s^{2})(5.50s)$ $ v_{y}=-17.5 m/s$ $\vec{v} $=$( 21.3m/s)\hat{i}+(-17.5 m/s)\hat{j}$ $|\vec{v} |=\sqrt{( 21.3m/s)^{2}+(-17.5 m/s)^{2}}=27.6$ m/s
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.