How to Determine Elevator Height at Time T1?

[Mr Davis 97]

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/s^2, and hits the ground at time $t = T_2$. Find the height of the elevator time $T_1$.

Homework Equations

SUVAT Equations

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?

 

[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.

 

[Mr Davis 97]

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

 

[ehild]

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?

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.

 

[Mr Davis 97]

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.

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...

 

[ehild]

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...

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

 

[Mr Davis 97]

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

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

 

[ehild]

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

 

[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.

 

[Hkj_1029]

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.

T2 is the time it took the ball to hit the ground (T2 is being measured through a different clock and it was started the instant ball stated falling through the floor.) not the final time when the event end (starting point being when elevator started to rise).

 

[malawi_glenn]

Find the height of the elevator time T1.

height of a time?

 

[haruspex]

T2 is the time it took the ball to hit the ground (T2 is being measured through a different clock and it was started the instant ball stated falling through the floor.) not the final time when the event end (starting point being when elevator started to rise).

Not so. It is clearly stated that

At t = 0, an elevator departs from the ground … marble falls … and hits the ground at time $t = T_2$.

Anyway, the thread is six years old.