Thread: Elevator

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

So we use the following equations:

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

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

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

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

Elevator

$\displaystyle v_y = v_{y0}- gT_1 $

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

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

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

Am I on the right track?

Hello, tukeywilliams!

It's easier than you think . . .

An elevator ascends from the ground with uniform speed.
At time $\displaystyle T_1 $ a boy drops a marble through the floor.
The marble falls with uniform acceleration $\displaystyle g = 9.8$ m/s²
. . and hits the ground $\displaystyle T_2 $ seconds later.
Find the height of the elevator at time $\displaystyle T_1 $.

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

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


The problem begins when the marble is dropped.

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

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


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

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

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

Oh ok. Just had a mental block.

Thanks