## Fundamentals of Physics Extended (10th Edition)

$H_{2} > H_{1}$, because as time passes, the ball gains more and more vertical velocity so the vertical distance in the second half of distance will be greater than the distance traveled in the first half of distance.
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 consider $\frac{L}{2}$). First, we need to convert the units of velocity: $v_{0}=161\dfrac {km}{h}=161\dfrac {1000m}{3600s}\approx 44.72\dfrac {m}{s}$ So the time needed to travel the first half distance will be: $t_{1}=\dfrac {L}{2v_{0}}=\dfrac {18.3m}{2\times 44.72\dfrac {m}{s}}\approx 0.205s$ 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 traveled 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 ball traveled in the second half of distance will be: $H_{2}=H_{total}-H_{1}\approx 61.8cm$ So why is $H_{2} > H_{1}$? This is because as time passes, the ball gains more and more vertical velocity so the vertical distance in the second half of distance will be greater than the distance traveled in the first half of distance.