# Pitcher Throwing A Ball In Protest

1. Sep 11, 2009

### Warmacblu

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F = ma

A kinematic equation, not sure which one though.

3. 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: Sep 11, 2009
2. Sep 11, 2009

### Andrew Mason

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: Sep 11, 2009
3. Sep 11, 2009

### Warmacblu

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.

4. Sep 11, 2009

### Staff: Mentor

Look for a kinematic equation that relates velocity and distance.

5. Sep 11, 2009

### Warmacblu

I believe this is the one:

Vf2 = Vi2 + 2a (Xf - Xi)

6. Sep 11, 2009

### Staff: Mentor

That's the one you need. Assume that the ball starts from rest.

7. Sep 11, 2009

### Warmacblu

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.

8. Sep 11, 2009

### Staff: Mentor

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.

9. Sep 11, 2009

### Warmacblu

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. Sep 11, 2009

### Staff: Mentor

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.

11. Sep 11, 2009

### Warmacblu

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. Sep 11, 2009

### Staff: Mentor

Good. The net force on the ball is 6.5 N. (Since that force is upward, let's call it positive.)

OK, the weight is 1.65 N downward, so negative.

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. Sep 11, 2009

### Warmacblu

Okay, I understand now. Thanks for all the help.

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