Calculating Impulse Force for a Pitched Ball: How to Find the Final Velocity?

AI Thread Summary
To determine the impulse force on a baseball hit vertically after being pitched, the change in momentum must be calculated using initial and final velocities. The initial horizontal velocity is 38.45 m/s, while the final vertical velocity can be found using the maximum height of 44.196 m, yielding an initial vertical velocity of 29.43 m/s. The change in momentum is calculated by subtracting the initial momentum vector from the final momentum vector, which requires understanding vector addition and subtraction. The final impulse force, after dividing the change in momentum by the contact time of 3/1000 seconds, is found to be approximately 2286.57 N at an angle of 52.6 degrees west of north. Mastery of vector operations is crucial for solving such problems effectively.
lab-rat
Messages
44
Reaction score
0

Homework Statement


A fast ball is pitched at a batter who hits a pop fly vertically into the air. The ball was moving horizontally at 38.45 m/s when it was hit by the bat. As it left the bat it was moving vertically and reached a maximum height of 44.196m. The mass of a baseball is 0.1417 kg. If the ball was in contact with the bat for 3/1000 of a second, determine the magnitude and direction of the impulse force with which it was struck.


I need to find the change in momentum. So I need to find the initial and final velocity. The only thing left that I can't do is find the final velocity. How do I calculate it when all that I know is the vertical height?
 
Physics news on Phys.org
lab-rat said:

Homework Statement


A fast ball is pitched at a batter who hits a pop fly vertically into the air. The ball was moving horizontally at 38.45 m/s when it was hit by the bat. As it left the bat it was moving vertically and reached a maximum height of 44.196m. The mass of a baseball is 0.1417 kg. If the ball was in contact with the bat for 3/1000 of a second, determine the magnitude and direction of the impulse force with which it was struck.


I need to find the change in momentum. So I need to find the initial and final velocity. The only thing left that I can't do is find the final velocity. How do I calculate it when all that I know is the vertical height?

For vertical motion, there is a constant downwards acceleration of g [we are told to take g= 9.8, 9.81 or 10 depending what level Physics you are taking]

From that , and the height reached, you can calculate the initial velocity.
 
but I don't have the time and the formula is v=vo + at
 
or s=vot + a(1/2 a) t2
 
lab-rat said:
or s=vot + a(1/2 a) t2

ever heard of V2 = Vo2 + 2as
 
OK, but the acceleration would be negative, so how can I find the square root?
 
lab-rat said:
OK, but the acceleration would be negative, so how can I find the square root?

Yes, it is negative, but you have to move it to the other side of the equation [whence it will become positive] since you are trying to find Vo.

V = 0 because it has finally stopped going up when it reaches maximum height.
 
Ohh, I thought I was looking for V. Vo = 29.43 j m/s

So I found Pi= 5.448i kgm/s
and Pf= 4.17j kgm/s

I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?
 
lab-rat said:
Ohh, I thought I was looking for V. Vo = 29.43 j m/s

So I found Pi= 5.448i kgm/s
and Pf= 4.17j kgm/s

I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?

A pythagorus solution is what you are after.

The change in momentum has to get rid of a lot of horizontal momentum, and produce a lot of vertical momentum at the same time, so it will have an up component, and a "back towards the pitcher" component
 
  • #10
So Vo is 29.43 m/s, and the back towards pitcher component is -29.43 m/s correct?

And I would just need to find the horizontal velocity ?
 
  • #11
lab-rat said:
So Vo is 29.43 m/s, and the back towards pitcher component is -29.43 m/s correct?

And I would just need to find the horizontal velocity ?

No. The ball arrived at 38+ m/s, then went up at 29+ m/s, so the change is 38+ back, and 29+ up.
 
  • #12
OK, so what would the variation of momentum be ?

P = mv

I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.

I need either two horizontal vectors or two vertical vectors

Then I just need to divide the change in momentum by 3/1000
 
  • #13
OK so I used the pythagorea theorem but that gave me the resulting momentum when I only need the difference in momentum.

It's probably really simple, I'm just not seeing it right now...
 
  • #14
I ended up just doing (4.17j-5.448i) / 3/1000
Fnet=(1390j-1816i) N

does that work?
 
  • #15
lab-rat said:
I ended up just doing (4.17j-5.448i) / 3/1000
Fnet=(1390j-1816i) N

does that work?

NO.

This question is basically a test of whether you can add and subtract vectors. Looks like you can't.
It is done by drawing arrows which forms triangles which you use trigonometry and/or pythagorus to solve. Google search if you can't find it in your own text.
 
  • #16
lab-rat said:
OK, so what would the variation of momentum be ?

P = mv

I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.

I need either two horizontal vectors or two vertical vectors

Then I just need to divide the change in momentum by 3/1000

The bit I highlighted is a real problem. You have to learn how to do that.
 
  • #17
I actually do know how to subtract vectors, but my professor had said that we did not have time to cover that section, so we wouldn't see vector addition/subtraction.
So I assumed we wouldn't have any assignments on it, and thought there was another way to do this problem.

Anyways, I ended up doing it anyway and found 2286.57N (N52.6degresW)
 
Back
Top