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: 28c

Answer

$H=67.5$ m

Work Step by Step

At the highest point, $y=H$, the vertical component of velocity is zero ($v_{y}=0$ ) Use 4-24 to find y=H $v_{y}^{2}=(v_{0}\sin\theta_{0})^{2}-2g(y-y_{0})$ $0=(v_{0}\sin\theta_{0})^{2}-2gH$ $2gH=(v_{0}\sin\theta_{0})^{2}$ $H=\displaystyle \frac{(v_{0}\sin\theta_{0})^{2}}{2g}$ $H=\displaystyle \frac{(42.0m/s)\sin 60^{o}}{2(9.80m/s^{2})}$ $H=67.5$ m
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