Posted by ahhhh! on Wednesday, May 25, 2011 at 8:51pm.
height reached during acceleration first
h = (1/2) a t^2 = (1/2)(29.4)(81) = 1191 meters
now speed up at 1190 meters
Vi = at = 29.4*9 = 265 m/s
so
we have a rocket at 1190 meters going at 265 m/s and with acceleration g = -9.8 m/s^2
When will the speed up hit zero
0 = Vi -gt
t = 265/9.8 = 27 seconds more up
how high then
h = Hi + Vi t + (1/2) a t^2
h = 1190 + 265*27 -4.9*27^2
h = 4773 m
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