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Homework Help: Max Jumping Height in Elevator Descending at Constant Vel.

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    I work as a supplemental instructor for an intro physics class. As I was preparing my worksheet for this week I came across this problem.

    A basketball player can jump to a height of 55cm. How far above the floor can he jump in an elevator that is descending at a constant 1.0m/s.

    2. Relevant equations
    vf2 = vi2 + 2aΔx

    3. The attempt at a solution
    First I used the above equation to determine the initial velocity the basketball player would need to achieve a height of 55cm on steady ground: 3.28m/s.

    I then said that relative to the floor of the elevator the player is moving at 4.28m/s initially.

    Thus his max height should be 93cm.

    However, the answer in the back of the textbook says that the answer is 55cm (same as his normal jump height).

    So that makes sense if the elevator is already moving when he jumps. Then his velocity before the jump is already -1m/s, offsetting the fact that the elevator is moving away from him after he jumps.

    If the elevator begins moving the instant after he begins jumping, would my original solution be correct?
    Last edited: Sep 26, 2016
  2. jcsd
  3. Sep 26, 2016 #2


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    Gold Member

    The elevator is moving at constant velocity. If the 100kg occupant whips out a scale, he will weigh 100kg. He will not even know he is moving.

    The appropriate way to frame this question is to acknowledge that constant motion is relative. The -1m/s of the elevator (relative to some external reference point) is irrelevant to anything within the elevator. It should not be mentioned in the explanation at all.

    It is a conceptual question because no math is required (indeed, it is a red herring). What is required is to recognize that the motion of the elevator being constant means the answer is the same as if he were standing motionless on the Earth.

    The elevator accelerating is an entirely different question, and is not answerable without further details.
    Last edited: Sep 26, 2016
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