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Homework Help: Find the Height of the Elevator

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    An elevator rises at a constant velocity. A marble is dropped from a hole in the elevator at time [itex]T_1[/itex]. It accelerates downward at 9.8 m/s^2 and hits the floor at time [itex]T_2[/itex]. Define the height at time [itex]T_1[/itex]

    2. Relevant equations
    [itex]y=y_0+v_0 t-1/2gt^2[/itex]

    3. The attempt at a solution
    So this sounds easy as cake. The initial velocity is 0 and we can assume [itex]T_1[/itex] is 0. Ok, so I can define the function of position as:

    [itex]y=y_0-4.9t^2[/itex] which at time [itex]T_2[/itex] equals 0.



    Sound reasonable?

    But for some reason the hint in the back of the book is this example of what the answer is supposed to look like. It says:

    [itex]T_1=T_2=4[/itex] then [itex]y_0=39.2[/itex]

    Which doesn't line up with my equation that I've used to determine the height. So idk. What did I do wrong?
  2. jcsd
  3. Jan 10, 2015 #2


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    I don't understand. Are we dropping the marble through the floor of the elevator to find the height of the elevator above the ground? Or are we dropping the marble through the roof of the elevator to find how tall the elevator(-box) is? Or something else I am misunderstanding?

    The initial velocity of what is zero, the marble? Zero with respect to what?
  4. Jan 10, 2015 #3
    It's really not that detailed man. Consider the marble and elevator just two particles. And the marble has zero initial velocity because it's dropped down without any velocity being given to it by the kid dropping it.
  5. Jan 10, 2015 #4


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    Really? You are, presumably, using the floor that it lands on as your frame of reference. Do you really believe that the marble has zero initial velocity relative to that?
  6. Jan 10, 2015 #5
    I see what you're getting at.
    How's this to define the velocity:

    [itex]v=-v_0+gt[/itex] for [itex]t≥T_1[/itex]

    so the position is:


    at [itex]t=T_2[/itex] the height is 0, so:


    I get it. The signs are a bit funky for me but it's good. I sort of worked the velocity function to fit the results of the hint. But doesn't the initial velocity still have to be zero for this equation to hold and still fit the hint?
  7. Jan 10, 2015 #6


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    The problem statement says that the elevator is traveling upward at a constant velocity. RELATIVE TO WHAT? Do you understand that it is utterly meaningless to say "this thing is moving" unless you say what it is moving relative TO ?
  8. Jan 10, 2015 #7
    Lose the CAPS lock. Use italics instead. Has more class and doesn't come off as abrasive :D

    Edit: Added smiley so I don't sound abrasive.

    And yes, I do understand frames of reference enough to know what you're saying. But it's not stated in the book. So I just assumed it was from the elevator's frame of reference. And that seems to fit the results that agree with the hint in the back of the book. Thank you for the insight.
  9. Jan 10, 2015 #8


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    But you specifically say that the elevator is moving. Even though you don't say relative to what, it doesn't matter if you are then saying that the marble ISN"T moving. The marble and the elevator are together. Pick a frame of reference and stick with it. No matter WHAT frame of reference you choose, the elevator and the marble start off with the same velocity, as per the problem statement. That's what I've been trying to get you to see.

    As for the caps, we all like to do what we're best at and abrasive is one of my best things :smile:
  10. Jan 10, 2015 #9
    So just to be sure I'm not silly and completely missing your point:

    If the FoR is the elevator the initial velocity of the marble when it drops is 0 because it's moving at a constant velocity, but only moving relative to the floor. Not to me. (If my understanding of SR hasn't failed me).

    I think I see a better way to get the equation now. If my FoR is the floor, and I build a position function for the marble from the time the elevator leaves the ground to the time the marble hits the floor I get

    [itex]y=v_0-9.8t[/itex] for [itex]t<T_1[/itex]

    [itex]y=y_0-4.9t^2[/itex] for [itex]T_1≤t≤T_2[/itex]

    To find the height I set these equal to eachother, and solve for [itex]y_0[/itex] to get:


    so my answer is:


    Is my understanding solid. Or is there something subtle I'm missing?
  11. Jan 10, 2015 #10


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    No, if the FoR is the elevator, the initial velocity of the marble is 0 because it is NOT moving at a constant velocity or any other velocity. In the FoR of the elevator, the elevator is stationary and the marble is stationary (until released). In the FoR of the floor, both are traveling upwards at the same velocity at the point when the marble is released.
    Last edited: Jan 11, 2015
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