Momentum - Baseball Homework: Average Force, Direction

[tiggrulz13]

Homework Statement

A 0.145 kg baseball pitched horizontally at 39.0 m/s strikes a bat and is popped straight up to a height of 37.0 m. (a)If the contact time between bat and ball is 1.0 ms, calculate the magnitude of the average force between the ball and bat during contact. (b)Find the direction of the average force on the ball.

Homework Equations

p=mv
F=p/t

The Attempt at a Solution

This seemed like a really simple problem to me, but Mastering Physics said my answer was wrong. I tried to find momentum using p=mv, p= 0.145 kg(39.0 m/s) = 5.655
Then I plugged this into F=p/t and got F=5.655/0.001 s =5655N, which it said was wrong. Am I missing something? What do I do with the height?
Also I don't know what to do with part b at all.

 

[LowlyPion]

Homework Statement

A 0.145 kg baseball pitched horizontally at 39.0 m/s strikes a bat and is popped straight up to a height of 37.0 m. (a)If the contact time between bat and ball is 1.0 ms, calculate the magnitude of the average force between the ball and bat during contact. (b)Find the direction of the average force on the ball.

Homework Equations

p=mv
F=p/t

The Attempt at a Solution

This seemed like a really simple problem to me, but Mastering Physics said my answer was wrong. I tried to find momentum using p=mv, p= 0.145 kg(39.0 m/s) = 5.655
Then I plugged this into F=p/t and got F=5.655/0.001 s =5655N, which it said was wrong. Am I missing something? What do I do with the height?
Also I don't know what to do with part b at all.

The height should give you the initial velocity after leaving the bat shouldn't it?

 

[tiggrulz13]

How would you do that?

 

[LowlyPion]

How would you do that?

Look here:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[tiggrulz13]

Ok, I tried using x=xo + vot + 1/2at^2 to find the initial velocity after leaving the bat, but I got a huge number:
37=0 + Vo(0.001) + 1/2 (39/.001)(.001)^2
Vo=36980.5 m/s
I'm sorry, for some reason I'm just not getting it tonight.
I also thought about using conservation of energy to find V and got:
1/2mv^2 +mgh = 1/2mv^2 + mgh
1/2(.145)(39)^2 + 0 = 1/2(.145)V^2 + (.145)(9.8)(37)
and i got v=28.2099 m/s

 

[LowlyPion]

Ok, I tried using x=xo + vot + 1/2at^2 to find the initial velocity after leaving the bat, but I got a huge number:
37=0 + Vo(0.001) + 1/2 (39/.001)(.001)^2
Vo=36980.5 m/s
I'm sorry, for some reason I'm just not getting it tonight.
I also thought about using conservation of energy to find V and got:
1/2mv^2 +mgh = 1/2mv^2 + mgh
1/2(.145)(39)^2 + 0 = 1/2(.145)V^2 + (.145)(9.8)(37)
and i got v=28.2099 m/s

Consider using the V² = 2gh relationship to figure the vertical initial velocity. (At its height its V is 0.)

Now you have a system that had MV in the positive x direction get changed to MV in the positive up direction in .001 second.

 

[tiggrulz13]

I used the v^2 = 2gh equation to get v=26.9295
I then plugged this into the F=p/t equation:
F = 0.145(39-26.9295) / 0.001
F=1750.22
and that was wrong - and I don't have anymore attempts left on that part, apparently the correct answer was 6900 N.

For the second part I need to find the direction (theta) of the average force. How do I go about doing that?

(I'm sorry if I'm taking a lot of your time, thank you for trying to help me, for some reason physics is just not clicking with me tonight)

 

[LowlyPion]

I used the v^2 = 2gh equation to get v=26.9295
I then plugged this into the F=p/t equation:
F = 0.145(39-26.9295) / 0.001
F=1750.22
and that was wrong - and I don't have anymore attempts left on that part, apparently the correct answer was 6900 N.

For the second part I need to find the direction (theta) of the average force. How do I go about doing that?

(I'm sorry if I'm taking a lot of your time, thank you for trying to help me, for some reason physics is just not clicking with me tonight)

This is incorrect. Velocity is a vector.

This means that the change in velocity is a right triangle (horizontal pitch to vertical pop fly). The magnitude of the change in velocity then is given by the root sum of the squares of the two.
(26.93² + 39²)^1/2

Using that change in speed times the mass divided by the time is the average force applied.

As to angle then, it is the angle of the formed by the vectors of velocity - but against the direction of the initial pitch. That would then be the Tan of Vup/Vpitch above the direction the pitch arrived from.