Find Height of Elevator at T_1: K&K Q1.17

[Radarithm]

Homework Statement

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 $T_1$

Homework Equations

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

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.

 

[Radarithm]

edit: nevermind.

 

[jbunniii]

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 $T_1 + T_2$.

 

[HalfLostAndSearching]

It seems like you are on the right track with using the equations for position and velocity with respect to time. However, it's important to remember that the elevator is moving at a constant speed, so its velocity is not changing. This means that the initial velocity (v0) is equal to the final velocity (vf).

Using the first equation you provided, we can set the initial velocity (v0) to vf and solve for time (T1). This will give us the time at which the marble is dropped.

Then, using the third equation, we can find the initial height (y0) of the elevator by setting the position at T1 to be equal to the position of the marble when it is dropped (which is 0).

So, we have:

v0 = vf = g(T2 - T1)

y0 = g(T1^2)/2

Substituting the value of T1 from the first equation into the second equation, we get:

y0 = g[(gT2^2)/(2g^2)]

y0 = T2^2/2g

Therefore, the height of the elevator at T1 is T2^2/2g, which is equal to 39.2m when T1 = T2 = 4s.

I hope this helps clarify the solution for you. Remember to always check your units and make sure they are consistent throughout your calculations. Good luck with your studies!