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March 27, 2015

March 27, 2015

Posted by **Nick** on Monday, September 7, 2009 at 11:56pm.

- physics -
**drwls**, Tuesday, September 8, 2009 at 12:26amLet 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

- physics -
**amy**, Sunday, September 14, 2014 at 8:18pmTwo diamonds begin a free fall from rest from the same height 1.2 s apart. How long after the first object begins to fall will the two objects be 20 m apart?

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