Question

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g. A) Draw a free-body diagram of the ball during the pitch. The ball is moving to the right.
B) What force did the pitcher exert on the ball during this record-setting pitch?
C) Estimate the force in Part B as a fraction of the pitcher’s weight. Assume that the pitcher has a mass of 84 kg.

Answer 1

A) In attachment

Find the free-body diagram in attachment. There are only two forces acting on the ball during the pitch:

- The force applied by the pitcher, F, which acts in the horizontal direction

- The force of gravity acting on the ball, also called weight, whose magnitude is

$W=mg$

where m is the mass and g is the acceleration of gravity, and it acts in the downward direction.

B) 160.2 N

First of all, we need to find the acceleration of the ball. We can do it by using the SUVAT equation

$v^2-u^2=2ad$

where

v = 47 m/s is the final velocity of the ball

u = 0 is the initial velocity (the ball starts from rest)

a is the acceleration

d = 1.0 m is the distance covered by the ball while the force is applied

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 m/s^2$

And since we know the mass of the ball,

m = 145 g = 0.145 kg

We can find the force exerted on it by using Newton's second law:

$F=ma=(0.145)(1104.5)=160.2 N$

C) 0.195 times the pitcher's weight

The weight of the pitcher is given by

W = mg

where

m = 84 kg is the mass of the pitcher

g = 9.8 m/s^2 is the acceleration of gravity

Solving,

$W=(84)(9.8)=823.2 N$

So now we can express the force found in part B) as a fraction of the pitcher's weight:

$\frac{F}{W}=\frac{160.2}{823.3}=0.195$