# Marbles dropped from elevator. Find height of elevator.

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

$$v_y = v_{y0}$$

$$y = v_{y0}t$$

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

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

Am I on the right track?

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Also $$0 = v_{yT_{2}}(T_1 + T_2) + \frac{1}{2}g(T_1 + T_2)^{2}$$

PhanthomJay