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
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 29.4 . The acceleration period lasts for time 9.00 until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 .
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