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?