K&K Question 1.17 (1 Viewer)

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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]\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.
When the marble is dropped its initial velocity is the same as the elevator's velocity upwards.


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edit: nevermind.


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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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