Elevator

**tukeywilliams** · November 27th 2007, 10:05 AM

An elevator ascends from the ground with uniform speed. At time $T_1$ a boy drops a marble through the floor. The marble galls with uniform acceleration $g = 9.8$ and hits the ground $T_2$ seconds later. Find the height of the elevator at time $T_1$ .

So we use the following equations:

$v_{y} = v_{y0} -gt$
$y-y_0 = v_{y0}t - \frac{1}{2}gt^{2}$

$y-y_0 = \frac{1}{2}(v_{y0} + v_y)t$

$v_{y}^{2} = v_{y0}^{2} - 2g(y-y_0)$

So we probably have to consider 2 cases: the elevator and the marble.

Elevator

$v_y = v_{y0}- gT_1$

$y-y_0 = v_{y0}T_{1} - \frac{1}{2}gT_{1}^{2}$

$y = \frac{1}{2}v_{y}T_{1}$

$v_{y}^{2} = -2g(y-y_0)$

Am I on the right track?

**Soroban** · November 27th 2007, 10:44 AM

Hello, tukeywilliams!

It's easier than you think . . .

An elevator ascends from the ground with uniform speed.
At time $T_1$ a boy 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 $T_1$ .

The height of the marble is given by: . $y \;=\;h_o + v_ot - \frac{1}{2}gt^2$

. . where: . $\begin{Bmatrix}h_o & = & \text{initial height} \\ v_o &=& \text{initial velocity} \\ g& = & 9.8\text{ m/s}^2 \end{Bmatrix}$

The problem begins when the marble is dropped.

Hence: . $T_1 \:= \:0$
. . . . . . $h_o \:= \:\text{height of elevator}$
. . . . . . $v_o \:= \:\text{speed of elevator} \:=\: v_e$

We have: . $y \;=\;h_o + v_et - 4.9t^2$

We are told: .when $t = T_2,\;y \:=\:0$

So we have: . $h_o + v_e(T_2) - 4.9(T_2)^2 \;=\;0$

. . Therefore: . $\boxed{\:h_o \;=\;4.9(T_2)^2 - v_e(T_2)\:}$

**tukeywilliams** · November 27th 2007, 11:40 AM

Oh ok. Just had a mental block.

Thanks

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