Posted by Nick on Monday, September 7, 2009 at 11:56pm.
Let t=0 be the time the first one is dropped. The distance that it falls after that is
y1 = (g/2)t^2 = 4.9 t^2
The distance that the second one has fallen at time t(>0.9s) is
y2 = (g/2)(t-0.9)^2 = 4.9(t-0.9)^2
You want to solve for the time t when
y1 - y2 = 4.9 [t^2 - (t-1)^2] = 11 m
2t - 1 = 11/4.9 = 2.24 s
t = 1.62 s
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