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 86: 44d


$ 61.8cm$

Work Step by Step

First, lets calculate the time needed for the ball to travel the first half of given distance ( lets assume the given distance is $ L$ so in this case we need to find $\frac{L}{2}$) We need to convert the units of velocity using the data ($1hr=3600s;1km=1000m$). So, $v_{0}=161\dfrac {km}{h}=161\dfrac {1000m}{3600s}\approx 44.72\dfrac {m}{s}$ Therefore, the time needed to travel the first half distance will be: $t_{1}=\dfrac {L}{2v_{0}}=\dfrac {18.3m}{2\times 44,72\dfrac {m}{s}}\approx 0.205s$ Also, because there is no change in horizontal velocity, the time needed to travel the second half of the given distance will be the same as time needed to travel first half of given distance. So, $t_{2}=t_{1}=\dfrac {L}{2v_{0}}=\dfrac {18.3m}{2\times 44.72\dfrac {m}{s}}\approx 0.205s$ Therefore, the descent of the ball till the ball travels the given distance will be: $H_{total}=\dfrac {g\left( t_{1}+t_{2}\right) ^{2}}{2}\approx \dfrac {9.8\dfrac {m}{s^{2}}\left( 0.41s\right) ^{2}}{2}\approx 82.4cm$ Initially, the ball has no vertical velocity so the vertical distance the ball travels in that time will be: $H_{1}=\dfrac {gt_{1}^{2}}{2}=\dfrac {9.8\dfrac {m}{s^{2}}\left( 0.205s\right) ^{2}}{2}\approx 20.6cm$ So the vertical distance the ball traveled in second half will be: $H_{2}=H_{total}-H_{1}\approx 61.8cm$
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