Pitcher Throwing A Ball In Protest

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A baseball pitcher throws a ball upward at 11.4 m/s, moving his hand through 1.68 m, and the problem requires calculating the force he exerts on the ball. The relevant equations include F = ma and kinematic equations, specifically Vf² = Vi² + 2a(Xf - Xi). The correct approach involves calculating the acceleration needed to achieve the final velocity while considering the effects of gravity. The net force is determined by combining the force exerted by the pitcher and the gravitational force acting on the ball. The final calculations yield a force of approximately 6.5 N from the pitcher, leading to a total force exerted of around 8.15 N when gravity is accounted for.
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Homework Statement



As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 11.4 m/s. In the process, he moves his hand through a distance of 1.68 m.

Acceleration of gravity is 9.8 m/s2.

If the ball has a mass of 0.168 kg, find the force he exerts on the ball to give it this upward speed. Answer in units of N.

Homework Equations



F = ma

A kinematic equation, not sure which one though.

The Attempt at a Solution



F = ma
F = .168 kg * (11.4 / 1.68) = 1.14

F = ma
F = .168 kg * (1.68 / 11.4) = .025

I did look at old threads regarding this question but cannot understand the hints. I guess I need to find a proper acceleration taking into account all the forces acting upon the ball but I do not know which kinematic equation to use.

Any help is appreciated,
Thanks
 
Last edited:
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Warmacblu said:

Homework Statement



As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 11.4 m/s. In the process, he moves his hand through a distance of 1.68 m.
Try using an energy approach. What is the kinetic energy of the ball when it leaves the pitcher's hand? How does that relate to the force applied by the pitcher over the 1.68 m. distance? Be careful to include the force of gravity and the increase in potential energy over that distance.

AM
 
Last edited:
Andrew Mason said:
Try using an energy approach. What is the kinetic energy of the ball when it leaves the pitcher's hand? How does that relate to the force applied by the pitcher over the 1.68 m. distance? Be careful to include the force of gravity and the increase in potential energy over that distance.

AM

We haven't really discussed energy yet which could contribute to why I am having difficulty understanding this problem.

However, I looked ahead a bit and found that k = 1/2mv2.

I do not know if that v is initial velocity or not and I do not know how I can relate it to the force if I can't figure out an acceleration for the force equation.
 
Look for a kinematic equation that relates velocity and distance.
 
Doc Al said:
Look for a kinematic equation that relates velocity and distance.

I believe this is the one:

Vf2 = Vi2 + 2a (Xf - Xi)
 
Warmacblu said:
I believe this is the one:

Vf2 = Vi2 + 2a (Xf - Xi)
That's the one you need. Assume that the ball starts from rest.
 
Doc Al said:
That's the one you need. Assume that the ball starts from rest.

Okay, here's what I did:

Vf2 = Vi2 + 2a (Xf - Xi)

0 = 11.4 + 2a (1.68)
a = -11.4 / (2 * 1.68)
a = -3.39

I don't think I can just plug this into F = ma because I haven't taken gravity into consideration, but I don't know where to take that into account.
 
Warmacblu said:
Okay, here's what I did:

Vf2 = Vi2 + 2a (Xf - Xi)

0 = 11.4 + 2a (1.68)
a = -11.4 / (2 * 1.68)
a = -3.39
You forgot to square the velocity. And you have the wrong sign because you mixed up Vi and Vf. (Vi = 0.)

I don't think I can just plug this into F = ma because I haven't taken gravity into consideration, but I don't know where to take that into account.
Using F = ma will tell you the net force. You'll need to solve for the force the player exerts on the ball, taking into account that gravity also acts on the ball.
 
Doc Al said:
You forgot to square the velocity. And you have the wrong sign because you mixed up Vi and Vf. (Vi = 0.)


Using F = ma will tell you the net force. You'll need to solve for the force the player exerts on the ball, taking into account that gravity also acts on the ball.

Okay, here's what I have now:

Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
a = 38.68 + (-9.8)
a = 28.88

F = ma
F = .168 * 28.88
F = 4.85

How does that look? Is my math and gravity assumption correct?
 
  • #10
Warmacblu said:
Okay, here's what I have now:

Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
So far, so good. Solve for a.
a = 38.68 + (-9.8)
a = 28.88
Do not try to take shortcuts! Solve for a, find the net force, then worry about the force of gravity

Hint: Two forces act on the ball, which combine to give you the net force.
 
  • #11
Doc Al said:
So far, so good. Solve for a.

Do not try to take shortcuts! Solve for a, find the net force, then worry about the force of gravity

Hint: Two forces act on the ball, which combine to give you the net force.

Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
77.36 = 2a
a = 38.68

F = ma
F = .168 * 38.68
F = 6.50

Now, to worry about the force of gravity.

F = ma
F = .168 * (-9.8) = -1.65

So ...

6.50 + (-1.65) = 4.85

However, I am unsure if gravity should be negative or not:

F = ma
F = .168 * 9.8 = 1.65

So ...

6.50 + 1.65 = 8.15

I need a little bit of reinforcement (explanation) on what sign gravity should have in this problem.
 
  • #12
Warmacblu said:
Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
77.36 = 2a
a = 38.68

F = ma
F = .168 * 38.68
F = 6.50
Good. The net force on the ball is 6.5 N. (Since that force is upward, let's call it positive.)

Now, to worry about the force of gravity.

F = ma
F = .168 * (-9.8) = -1.65
OK, the weight is 1.65 N downward, so negative.

So ...

6.50 + (-1.65) = 4.85
Here's where you went wrong.

Think this way:
Net Force = Force of Pitcher (upward) + Force of gravity (downward)
+ 6.5 = Fp - 1.65

solve for Fp:
Fp = 6.5 + 1.65
 
  • #13
Okay, I understand now. Thanks for all the help.
 

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