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K&K Question 1.17

  1. Mar 23, 2014 #1

    Radarithm

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    1. The problem statement, all variables and given/known data
    At t = 0, an elevator departs from the ground with uniform speed. At time [itex]T_1[/itex] a child drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s2, and hits the ground [itex]T_2[/itex] seconds later. Find the height of the elevator at time [itex]T_1[/itex]


    2. Relevant equations

    [tex]\frac{dv}{dt}=g[/tex]
    [tex]\dot{y}=g\int dt[/tex]
    [tex]y=g\int_{T_1}^{T_2}t dt=g\left(\frac{T_2^2-T_1^2}{2}\right)+y_0[/tex]
    [tex]y=g\int_0^{T_2}t dt=\frac{gT_2^2}{2}+y_0[/tex]

    3. The attempt at a solution

    I was sure of the fact that if I set the 3rd equation above to equal zero, I could solve for the initial height; the problem seemed confusing at first but is actually quite trivial. I turned to the back of the book to look for the answer (and I was sure that I was correct), but I got a hint; if [itex]T_1=T_2=4 s[/itex] then [itex]h=39.2 m[/itex]
    The 3rd equation gave me zero, and the fourth one gave me 79.4 m or [itex]2h[/itex]. I was able to use this to solve for the velocity of the elevator, but that doesn't seem to help much. I'm not sure what I'm doing wrong; the height at [itex]T_1[/itex] is [itex]y_0[/itex]. Can someone help me out? This should have been so much easier. I've got to be making some mistake in the calculus.
     
  2. jcsd
  3. Mar 23, 2014 #2
    When the marble is dropped its initial velocity is the same as the elevator's velocity upwards.
     
  4. Mar 23, 2014 #3

    Radarithm

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    edit: nevermind.
     
  5. Mar 23, 2014 #4

    jbunniii

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    Be careful about the times. The marble is dropped at time ##T_1## and it hits the floor ##T_2## seconds later, i.e. at time ##T_1 + T_2##.
     
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