Projectile Motion-Baseball

  • Thread starter brita
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  • #1
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Homework Statement



At bat a baseball player hits a ball at a height of .889 m. The ball leaves the bat at 51 m/s at an angle of 81 degrees from the vertical. The ball skims the top of a 2.74 m wall as it leaves the field.
a.) Draw position, velocity and acceleration graphs describing the ball between the hit and when it leaves the field. Assume t=0 when the ball is hit, up is positive, the ball moves in positive horizontal direction, and that the pitcher's mound is the origin.
b) How long is the ball in the air?
c.) How far away from home plate is the wall?
d.) What angle relative to the x-axis is the ball moving when it passes the wall?
e.)If an infield player can jump to get his mitt 2.74 m off the ground, how close to home plate was he to still catch the ball.

Homework Equations



y=y0+v0sin(θ)t-1/2gt2
Δx=v0cos(θ)t
For the vector components:vy=v0sin(θ)-gt
vx=v0cos(θ)
x=x0+v0xΔt


The Attempt at a Solution


I've drawn the graphs for part a and I am sure I have done that correctly.

I used y=y0+v0sin(θ)t-1/2gt2 to find t which I calculated to be 10.326 s

I am confused on part C, ;the only reasonable equation seems to be Δx=v0cos(θ)t. But if I use this equation I will need an x(final) and x(initial). I know I am looking for x(final) and I think x(initial) is -18.29. But then I will have to add this to the answer for (v0cos(θ)t). But this does not make sense to me because I think I've already assumed 18.29 to be my origin and I've used this to calculate the time already. SO adding the 18.29 seems, to me, to be redundant.

For part e: I got an angle of 80.9 deg. And I know that should be slightly less than the launch angle. But I thought it would be a larger difference.

For part f: I am not sure what to do at all because I am not sure what the question is asking.

Thank you for your help.
 
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Answers and Replies

  • #2
Simon Bridge
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Welcome to PF;
I've drawn the graphs for part a and I am sure I have done that correctly.
Very good - getting these right will make everything else much easier.

I used y=y0+v0sin(θ)t-1/2gt2 to find t which I calculated to be 10.326 s
What did you use for the final height?
i.e. if you used the height of the wall - did the ball skim the top of the wall while it was on it;s way up or on it;s way down? Or was that right at the apex of the flight?

I am confused on part C, ;the only reasonable equation seems to be Δx=v0cos(θ)t. But if I use this equation I will need an x(final) and x(initial).
Do you know how far the pitcher's mound (the origin) is from the batsman's position (home plate)?
Note: you should be deriving the equations from the graphs rather than trying to pick them from a list. You said you drew these correctly - so you must have put the initial value of x on one of them.

I know I am looking for x(final) and I think x(initial) is -18.29. But then I will have to add this to the answer for (v0cos(θ)t). But this does not make sense to me because I think I've already assumed 18.29 to be my origin and I've used this to calculate the time already. SO adding the 18.29 seems, to me, to be redundant.
x=18.29 (units?) cannot be the origin. By definition, the origin is at x=0.

For part e: I got an angle of 80.9 deg. And I know that should be slightly less than the launch angle. But I thought it would be a larger difference.
The wall is somewhat higher than the height the ball got hit - but the parabola is a very steep one.
Consider - if the ball was hit straight up, then it would have the same angle all the way down as all the way up.

For part f: I am not sure what to do at all because I am not sure what the question is asking.
It's asking where the player has to be to catch the ball.
 
  • #3
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Thank you for the quick reply.
I used -2.473 as the final height since the ball hit the wall onthe way down. But I'm wondering if I should subtract the original height from this (so 2.473-.889) because the ball y(i) is not 0 but 0.889.

I know the pitcher's mound is 18.29 m away from the launch point. So since the origin is 0( which is the pitcher's mound), the position of the launch point is -18.29m. So I should add this to the distance I've found, because I've only calculated the distance from 0 to the final height and not from the point of launch. Right?

I still don't understand part e. If it's a steep parabola, then the difference between two angles in the parabola is not very large. Is that what you mean?

For part f: So I should calculate the distance to the point where the ball is 2.743m above the ground as it is still going up (before reaching max height)?
 
  • #4
462chevelle
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as far as c goes, your x velocity will be constant throughout the whole range of motion. so if you find your x component of velocity, you can just multiply that by time to get your distance.
 
  • #5
Simon Bridge
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Thank you for the quick reply.
I used -2.473 as the final height since the ball hit the wall onthe way down. But I'm wondering if I should subtract the original height from this (so 2.473-.889) because the ball y(i) is not 0 but 0.889.
Certainly the vertical displacement is ##\Delta y = y_f-y_i##

I know the pitcher's mound is 18.29 m away from the launch point. So since the origin is 0( which is the pitcher's mound), the position of the launch point is -18.29m. So I should add this to the distance I've found, because I've only calculated the distance from 0 to the final height and not from the point of launch. Right?
Try just redoing the calculation from scratch - using the correct initial horizontal position.

I still don't understand part e. If it's a steep parabola, then the difference between two angles in the parabola is not very large. Is that what you mean?
For an inverted parabola, the angle of approach does not change very much near the roots (the endpoints of the trajectory). Sketch a bunch and see.
If the parabola is tall and narrow like this one, then the height can change a lot without changing the angle or horizontal position much ...so a big difference in height does not make a big difference to the angle. Since the angle is the inverse-tan of the slope - it is easier to see in terms of gradients.
Again - sketch a bunch of them and see.

For part f: So I should calculate the distance to the point where the ball is 2.743m above the ground as it is still going up (before reaching max height)?
... that's what you have to decide. What makes sense? What happens if you do it the other way (while the ball is coming down) - where does the player have to stand?
 
  • #6
462chevelle
Gold Member
305
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Certainly the vertical displacement is ##\Delta y = y_f-y_i##

Try just redoing the calculation from scratch - using the correct initial horizontal position.

For an inverted parabola, the angle of approach does not change very much near the roots (the endpoints of the trajectory). Sketch a bunch and see.
If the parabola is tall and narrow like this one, then the height can change a lot without changing the angle or horizontal position much ...so a big difference in height does not make a big difference to the angle. Since the angle is the inverse-tan of the slope - it is easier to see in terms of gradients.
Again - sketch a bunch of them and see.

... that's what you have to decide. What makes sense? What happens if you do it the other way (while the ball is coming down) - where does the player have to stand?
wasn't the angle 81 degrees from the vertical? so it would be 9 degrees from horizontal, very long and short?
 
  • #7
Simon Bridge
Science Advisor
Homework Helper
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wasn't the angle 81 degrees from the vertical? so it would be 9 degrees from horizontal, very long and short?
Oh did I misread that - in which case, switch "vertical" for "horizontal" etc and the statement still stands.
The ends near the roots don't change their slope by much.

Note: in the formulas in post #1, ##\theta## is the angle from the horizontal ... another source of possible mistakes for brita.
 

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