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
[tex]W=mg[/tex]
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
[tex]v^2-u^2=2ad[/tex]
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,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 m/s^2[/tex]
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:
[tex]F=ma=(0.145)(1104.5)=160.2 N[/tex]
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,
[tex]W=(84)(9.8)=823.2 N[/tex]
So now we can express the force found in part B) as a fraction of the pitcher's weight:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
