Posted by daniel on Friday, June 15, 2012 at 7:48pm.
they both go at the same horizontal speed forever and ever (until they hit)
Therefore we only have to solve a dropping problem
h and 4 h and we want the times in the air
h = (1/2) g t^2
4 h = (1/2) g T^2
t^2 = 2 h/g
T^2 = 8 h/g
T^2/t^2 = 4
so
T/t = 2
so the high one is in the air twice as long
so
It goes twice as far.
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