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Relative motion in an elevator

  1. Jul 5, 2015 #1
    1. The problem statement, all variables and given/known data
    So I am in an elevator throwing a ball in an air, straight up and down, about 1 meter from my hand. The elevator is moving upwards at a rate of 2 m/s. From an observer on the ground, would they see the ball reach its peak at the same instant that I do? If not what is this difference in time.

    From my perspective, I see the ball move up 1 meter, then down 1 meter. How far up and down would the observer see the ball move?

    2. The attempt at a solution
    My thoughts are that the observer would see the ball go higher up, as opposed to 1 m. But then they would see the ball move a shorter distance on the way down. As for the first part, I'm not sure where to begin. My guess is that the times would not be the same, but again, I'm in a stump. Thanks
     
  2. jcsd
  3. Jul 5, 2015 #2
    I think you guessed right but maybe you could prove it using kinematics?
     
  4. Jul 5, 2015 #3
    Unless you are considering special relativistic effects, the time would be the same for both observer.
     
  5. Jul 5, 2015 #4

    haruspex

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    That would be true if they agreed about the event, but as the OP noted, they would disagree about the vertical speed of the ball at any given instant. In particular they would disagree about whether its vertical speed is zero.
     
  6. Jul 6, 2015 #5
    Why don't you do the math, and let the math answer the question for you?

    Chet
     
  7. Jul 6, 2015 #6
    I never considered this. It seems quite dramatic. I did some calculations and came up with the following general forms.

    From inside the elevator time of object to rise and fall:

    ##t=\frac { 2v_o}{g }##

    From outside the elevator:

    ##t=\frac{v_o+\sqrt { v_o^2-v_E^2 } +v_E}{g}##

    Although I didn't originate the question, I learned a lot. Thanks
     
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