# Elevator uniform speed problem

1. Sep 5, 2016

### Mr Davis 97

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. Sep 5, 2016

### J Hann

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.

3. Sep 5, 2016

### Mr Davis 97

I'm not seeing how I would find t...

4. Sep 5, 2016

### ehild

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.

5. Sep 5, 2016

### Mr Davis 97

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...

6. Sep 5, 2016

### ehild

It is the solution. Both T1 and T2 are given data, and h is the elevator time at T1.

7. Sep 5, 2016

### Mr Davis 97

Ah! How did I miss that... Thanks!

8. Sep 5, 2016

9. Sep 8, 2016

### thavab

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.