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Marbles dropped from elevator. Find height of elevator.

  1. Nov 27, 2007 #1
    An elevator ascends from the ground with uniform speed. At time [tex] T_1 [/tex] a boy drops a marble through the floor. The marble galls with uniform acceleration [tex] g = 9.8 [/tex] and hits the ground [tex] T_2 [/tex] seconds later. Find the height of the elevator at time [tex] T_1 [/tex].

    So we use the following equations:

    [tex] v_{y} = v_{y0} -gt [/tex]
    [tex] y-y_0 = v_{y0}t - \frac{1}{2}gt^{2} [/tex]

    [tex] y-y_0 = \frac{1}{2}(v_{y0} + v_y)t [/tex]

    [tex] v_{y}^{2} = v_{y0}^{2} - 2g(y-y_0) [/tex]

    So we probably have to consider 2 cases: the elevator and the marble.


    [tex] v_y = v_{y0} [/tex]

    [tex] y = v_{y0}t [/tex]

    [tex] y = \frac{1}{2}v_{y}t [/tex]

    [tex] v_{y}^{2} = -2g(y-y_0) [/tex]

    Am I on the right track?
  2. jcsd
  3. Nov 27, 2007 #2
    Also [tex] 0 = v_{yT_{2}}(T_1 + T_2) + \frac{1}{2}g(T_1 + T_2)^{2} [/tex]
  4. Nov 27, 2007 #3


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    Lots of equations there, you must choose the best one. If the marble is dropped at T1, it hits the ground in (T2-T1) seconds. You must now solve for the initial height of the elevator at the time the marble was dropped. That's also the initial height of the marble. Now once you know the initial velocity of the marble, which is related to the velocity of the elevator, which equation would you choose? Watch your plus and minus signs.

    EDIT: Sorry, I misread the problem statemnt, it takes T2 seconds to reach the ground, so just assume the start time T1 is 0, and use T2 in your kinematic equation that relates displacemnt with initial velocity, acceleration of gravity, and time.
    Last edited: Nov 27, 2007
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