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 87: 47a


$H=1,22+97,5-\dfrac {\left( 97,5\right) ^{2}}{107}\approx 9,88m$ Since the height of the fence is $H_{fence}=7,32m\lt H=9.88m$, the ball will clear the fence.

Work Step by Step

Lets calculate the height of the ball relative to the ground at given time $H_{t}=H_{0}+v_{0}t\sin \alpha -\dfrac {gt^{2}}{2}\left( 1\right) $ and since horizontal component of velocity of the ball doesn't change we can calculate the horizontal distance ball traveled at any given time by $X\left( t\right) =v_{0}t\cos \alpha \left( 2\right) $ So from (1) an (2) we get $H\left( X\right) =X\tan \alpha -\dfrac {gX^{2}}{2v^{2}_{0}\cos ^{2}\alpha } +H_{0}(3)$ İn this problem $H_{0}=1.22m$ ,$α=45^{0}$ So if we simplify 3 we will get $H\left( X\right) =1.22m+X-\dfrac {gX^{2}}{v^{2}_{0}}$ On the other hand we know the horizontal range of ball so we can calculate its initial velocity by $\dfrac {v^{2}_{0}\sin _{2}\alpha }{g}=X_{\max }=107m\Rightarrow v^{2}_{0}=\dfrac {X_{\max }g}{\sin 2\alpha }(4)$ So from (4) and (3) we get $H\left( X\right) =1,22+X-\dfrac {X^{2}}{X_{\max }}$ fence is located $97.5m $ away from the initial location of the ball horizontally ($X=97.5m ,X_{max}=107m$) so we get $H=1,22+97,5-\dfrac {\left( 97,5\right) ^{2}}{107}\approx 9,88m$ Since the Height of the fence is $H_{fence}=7,32m\lt H=9.88m$ The ball will clear the fence
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