- #1
tronter
- 185
- 1
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].
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?
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?