Pitcher Throwing A Ball In Protest

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Homework Help Overview

The problem involves a baseball pitcher throwing a ball straight up into the air with an initial speed of 11.4 m/s over a distance of 1.68 m. The discussion centers around calculating the force exerted by the pitcher on the ball, considering the effects of gravity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore both kinematic and energy approaches to determine the force exerted on the ball. Some suggest using the kinetic energy of the ball and its relationship to the force applied over the distance. Others discuss the need to account for gravitational force when calculating net force.

Discussion Status

The discussion is active with participants sharing various attempts at calculations and questioning the assumptions made regarding initial and final velocities. There is ongoing clarification about the role of gravity in the calculations, and some participants express uncertainty about the signs used in force equations.

Contextual Notes

Participants note that they have not yet discussed energy concepts in detail, which may contribute to their difficulties in understanding the problem. There is also mention of needing to find a proper acceleration and the appropriate kinematic equation to use.

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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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