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Free Fall - Calculating Origin/Release

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  1. Jan 28, 2015 #1
    1. The problem statement, all variables and given/known data
    In this problem I don't know the release point/origin of a dropped ball. I do however have a stroboscopic photograph of it moving through my frame of reference.
    In my frame of reference it moves 2.0m in 4 seconds. Now, I am suppose to figure out how far above the first point (point A) the ball was dropped. I know the acceleration is 0.01m/s^2.

    2. Relevant equations
    d = 1/2*a*t^2 + Vi*t + Xi

    3. The attempt at a solution
    Can I just solve out for Xi using d equal to 2.0m?
    2.0m - 1/2*0.01m/s^2*4^2 = Xi
     
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  3. Jan 28, 2015 #2

    PeroK

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    That's not going to work, as you don't know the velocity at either point.

    Can you think of a way to generate two equations using that equation of motion?
     
  4. Jan 28, 2015 #3
    I know the instantaneous velocity at two different points, but I am unsure of how that will help me. So, in theory I could plug stuff into Vf^2 = Vi^2 + 2*a*d, but that will only give me the distance between the two velocities. Which I know already. I also know that I can use Vf = Vi + a*t again I don't see how that will help me out.
     
  5. Jan 28, 2015 #4

    PeroK

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    One of us is misunderstanding the problem. How do you know the velocity at the two points? I thought all you knew was that it took 4 seconds to move 2m? You don't know the velocity anywhere except at the start (that's a hint!).
     
  6. Jan 28, 2015 #5
    Oh, I'm sorry. I have five points (A, B, C, D, and E). I know that between each point is 1 second. So, I have measured between each point and used the midpoint method to find the instantaneous velocity at B and D. I know the initial velocity is 0.0m/s, since it is a free fall.
     
  7. Jan 28, 2015 #6

    PeroK

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    So, what's the velocity at A?
     
  8. Jan 28, 2015 #7
    Wait... I think I might get it. If I know the instantaneous velocity on point B, and I know the instantaneous velocity at release (Vi = 0.0m/s) I can use Vf^2 = Vi^2 + 2*a*d. Then all I have to do is minus the distance from A to B to get the distance from release to A!
     
  9. Jan 28, 2015 #8

    PeroK

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    That's right. IF you know the velocity at a point you can find d that way. How are you going to find the velocity at B (or anywhere else)?
     
  10. Jan 28, 2015 #9
    I know that the instantaneous velocity at point B is equal to the average velocity from point A to point C (considering each point is 1 second apart). Sincethe average velocity between to points is equal to the instantaneous velocity at the midpoint. But, that is the only way I know how to do it.
     
  11. Jan 28, 2015 #10

    PeroK

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    Sounds good!
     
  12. Jan 29, 2015 #11
    If you have: a (0.01) s (2.0) and t (4.0)
    Use: u = ( s - ( ½ * a * t ² ) ) / t
    To find the intial velocity, then use as the final velocity (v) from the start in the following:
    Use: s = ( v ² ) / ( 2 * a )
     
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