Asked by S kumar
A stone is dropped from edge of a roof and is observed to pass a window, vertically below it. The stone takes 0.1 sec to fall between the top and bottom of the window. If the window is of 2 m height, how far is the roof above the top of the window?
Answers
Answered by
Henry
h = V*t + 0.5g*t^2 = 2 m.
V*0.1 + 4.9*0.1^2 = 2.
0.1V = 2-0.049 = 1.95.
V = 19.5 m/s at top of the window.
V^2 = Vo^2 + 2g*h.
V = 19.5 m/s, Vo = 0, g = 9.8 m/s^2, h = ?.
V*0.1 + 4.9*0.1^2 = 2.
0.1V = 2-0.049 = 1.95.
V = 19.5 m/s at top of the window.
V^2 = Vo^2 + 2g*h.
V = 19.5 m/s, Vo = 0, g = 9.8 m/s^2, h = ?.
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