Question

A pitcher throws a baseball horizontally from the mound to home plate. The ball falls 0.913 m (2.99 ft) by the time it reaches home plate 18.3 m (60 ft) away. How fast was the pitcher's pitch?

Answer 1

Answer:

42.42 m/s

Explanation:

Let g = 9.81 m/s2. The time it takes for the ball to fall vertically 0.913m

$s = gt^2/2$

$t^2 = 2s/g = 2*0.913/9.81 = 0.186$

$t = \sqrt{0.186} = 0.431 s$

This is also the time it takes for the ball to travel horizontally 18.3 m across. So the ball horizontal velocity is

18.3 / 0.431 = 42.42 m/s