# 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 $T_1$ a child drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s2, and hits the ground $T_2$ seconds later. Find the height of the elevator at time $T_1$

2. Relevant equations

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

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 $T_1=T_2=4 s$ then $h=39.2 m$
The 3rd equation gave me zero, and the fourth one gave me 79.4 m or $2h$. 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 $T_1$ is $y_0$. Can someone help me out? This should have been so much easier. I've got to be making some mistake in the calculus.

#### wakefield

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

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edit: nevermind.

#### jbunniii

Homework Helper
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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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