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

Answer 2

The velocity of the pitcher's pitch if the pitcher throws a baseball horizontally from the mound to home plate. is 42.22m/s

First, we need to get the time taken by the baseball to reach the home plate expressed using Newton's law of motion:

$S =ut+\frac{1}{2}gt^2\\0.913=0+ \frac{1}{2}(9.8)t^2\\0.913=4.9t^2\\t^2=\frac{0.913}{4.9}\\t^2= 0.186\\t = \sqrt{0.186}\\t= 0.4312s$

Get the velocity of the pitcher's pitch. Using the formula:

$v =\frac{s}{t}$

$v=\frac{18.3}{0.4312}\\v = 42.44m/s$

Hence the velocity of the pitcher's pitch if the pitcher throws a baseball horizontally from the mound to home plate. is 42.22m/s

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