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Elevator uniform speed problem

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    At t = 0, an elevator departs from the ground with uniform speed. At time ##T_1## a child drops a marble through the floor.The marble falls with uniform acceleration g = 9.8 m/s^2, and hits the ground at time ##t = T_2##. Find the height of the elevator time ##T_1##.

    2. Relevant equations
    SUVAT Equations

    3. The attempt at a solution
    Using the equation ##h = v_e T_1## and ##0 = h + v_e T_2 - \frac{1}{2} g T_2^2##, I was able to find the height in terms of ##T_1## and ##T_2##: ##\displaystyle h = \frac{g T_2^2}{2(1 + \frac{T_2}{T_1})}##

    However, this gives the height in terms of ##T_1## and ##T_2##, but I need it in terms of ##T_1##. How should I go about doing that?
     
  2. jcsd
  3. Sep 5, 2016 #2
    You need to find time t for the object to reach the ground after being dropped.
    t is just the difference between T1 and T2.
     
  4. Sep 5, 2016 #3
    I'm not seeing how I would find t...
     
  5. Sep 5, 2016 #4

    ehild

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    T1 and T2 are given: T1 is the instant of time when the child drops the marble and T2 is the time when it reaches the ground. The time of fall is T2-T1. You have to use the time of fall T2-T1 in the second equation.
     
  6. Sep 5, 2016 #5
    So now I have ##\displaystyle h = \frac{gT_1(T_2- T_1)^2}{2T_2}##, but this doesn't seem to really get me anywhere...
     
  7. Sep 5, 2016 #6

    ehild

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    It is the solution. Both T1 and T2 are given data, and h is the elevator time at T1.
     
  8. Sep 5, 2016 #7
    Ah! How did I miss that... Thanks!
     
  9. Sep 5, 2016 #8

    ehild

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    :smile:
     
  10. Sep 8, 2016 #9
    s=ut+1/2at^2
    h=0+1/2*9.8*(T2-T1)^2
    initial velocity of the ball related to the elevator is zero.
     
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