An elevator ascends from the ground with uniform speed. At time [tex] T_1 [/tex] a boy drops a marble through the floor. The marble galls with uniform acceleration [tex] g = 9.8 [/tex] and hits the ground [tex] T_2 [/tex] seconds later. Find the height of the elevator at time [tex] T_1 [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

So we use the following equations:

[tex] v_{y} = v_{y0} -gt [/tex]

[tex] y-y_0 = v_{y0}t - \frac{1}{2}gt^{2} [/tex]

[tex] y-y_0 = \frac{1}{2}(v_{y0} + v_y)t [/tex]

[tex] v_{y}^{2} = v_{y0}^{2} - 2g(y-y_0) [/tex]

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

Elevator

[tex] v_y = v_{y0} [/tex]

[tex] y = v_{y0}t [/tex]

[tex] y = \frac{1}{2}v_{y}t [/tex]

[tex] v_{y}^{2} = -2g(y-y_0) [/tex]

Am I on the right track?

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# Homework Help: Marbles dropped from elevator. Find height of elevator.

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