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Math Help - Elevator

  1. #1
    Senior Member tukeywilliams's Avatar
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    Elevator

    An elevator ascends from the ground with uniform speed. At time  T_1 a boy drops a marble through the floor. The marble galls with uniform acceleration  g = 9.8 and hits the ground  T_2 seconds later. Find the height of the elevator at time  T_1 .

    So we use the following equations:

     v_{y} = v_{y0} -gt
     y-y_0 = v_{y0}t - \frac{1}{2}gt^{2}

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

     v_{y}^{2} = v_{y0}^{2} - 2g(y-y_0)

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

    Elevator

     v_y = v_{y0}- gT_1

     y-y_0 = v_{y0}T_{1} - \frac{1}{2}gT_{1}^{2}

     y = \frac{1}{2}v_{y}T_{1}

     v_{y}^{2} = -2g(y-y_0)

    Am I on the right track?
    Last edited by tukeywilliams; November 27th 2007 at 09:18 AM.
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  2. #2
    Super Member

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    Hello, tukeywilliams!

    It's easier than you think . . .


    An elevator ascends from the ground with uniform speed.
    At time  T_1 a boy drops a marble through the floor.
    The marble falls with uniform acceleration  g = 9.8 m/sē
    . . and hits the ground  T_2 seconds later.
    Find the height of the elevator at time  T_1 .
    The height of the marble is given by: . y \;=\;h_o + v_ot - \frac{1}{2}gt^2

    . . where: . \begin{Bmatrix}h_o & = & \text{initial height} \\ v_o &=& \text{initial velocity} \\ g& = & 9.8\text{ m/s}^2 \end{Bmatrix}


    The problem begins when the marble is dropped.

    Hence: . T_1 \:= \:0
    . . . . . . h_o \:= \:\text{height of elevator}
    . . . . . . v_o \:= \:\text{speed of elevator} \:=\: v_e

    We have: . y \;=\;h_o + v_et - 4.9t^2


    We are told: .when t = T_2,\;y \:=\:0

    So we have: . h_o + v_e(T_2) - 4.9(T_2)^2 \;=\;0

    . . Therefore: . \boxed{\:h_o \;=\;4.9(T_2)^2 - v_e(T_2)\:}

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  3. #3
    Senior Member tukeywilliams's Avatar
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    Oh ok. Just had a mental block.

    Thanks
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