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?