## Fundamentals of Physics Extended (10th Edition)

$0.205$s
First, we need to convert $161\dfrac {km}{h} to \dfrac {m}{s}$. İn order to do that: we need $1hr=3600s;1km=1000m$ Therefore, $v_{0}=161\dfrac {km}{h}=161\dfrac {1000m}{3600s}\approx 44.72\dfrac {m}{s}$ This is the initial horizontal velocity of the ball and its horizontal velocity doesn't change since there is no horizontal force acting on the ball. So the time needed to travel half of this given distance will be (for the second half of distance, lets assume the distance is $L$ so the time required is needed to travel $\frac{L}{2}$): $t_{2}=\dfrac {L}{2v_{0}}=\dfrac {18.3m}{2\times 44.72\dfrac {m}{s}}\approx 0.205s$ The key here is that the horizontal velocity of the ball doesn't change.