# Question #a790e

Sep 26, 2017

The ball will fall 0.757 m while traveling from the pitcher to the catcher if thrown at 43.3 m/s.

#### Explanation:

What is the time to reach the catcher?
speed = distance/time so
time = distance/speed

$t = \frac{s}{v} = \frac{17.0 \cancel{m}}{43.3 \left(\frac{\cancel{m}}{s}\right)} = 0.393 s$

How far vertically would a ball fall, if just dropped, in 0.393 s?
$s = u \cdot t + \left(\frac{1}{2}\right) \cdot a \cdot {t}^{2}$
$s = u \cdot t + \left(\frac{1}{2}\right) \cdot g \cdot {t}^{2}$
$s = 0 \cdot 0.393 s + \left(\frac{1}{2}\right) \cdot 9.8 \frac{m}{s} ^ 2 \cdot {\left(0.393 s\right)}^{2} = 0.757 m$

The ball would also fall 0.757 m while traveling from the pitcher to the catcher if thrown at 43.3 m/s. If thrown at a speed in excess of 43.3 m/s, it will fall less distance.

I hope this helps,
Steve