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: 42a

Answer

$5.33$ m

Work Step by Step

At any given time, the height of the particle will be: $H_{t}=H_{0}+v_{0}\sin \theta _{0}t-\dfrac {gt^{2}}{2}\left.....( 1\right) $ where $H_{0}$ is initial height of particle, $v_{0}$ is initial velocity, $\theta_{0}$ is the initial angle of particle relative to horizontal line and $t$ is the time passed since object launched. At any given time, the horizontal distance traveled by the object will be $x=v_{0}\cos \theta _{0}t\left.......( 2\right) $ So using (1) and (2), we get $t=\dfrac {x}{v_{0}\cos \theta _{0}}\Rightarrow H_{t}=H_{0}+x\tan \theta _{0}-\dfrac {gx^{2}}{2v^{2}_{0}\cos ^{2}\theta _{0}} $ For the first wheel, $v_{0}=26,5\dfrac {m}{s};X=23m;\theta _{0}=53^{0};H_{0}=3m$ Substituting these values in the formula above, we get $H_{1}=23.33m$ The height of the wheel is $18$ m so, $\Delta H=H_{1}-H_{wheel}=23.33-18=5.33m$
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