posted by Apple on .
A rocket moves upward, starting from rest with an acceleration of 32.0 m/s2 for 3.27 s. It runs out of fuel at the end of the 3.27 s but does not stop. How high does it rise above the ground?
work this in two parts.
launch vf(at end of burn)=a*t
heightatendofburn;1/2 a t^2
Now the second part, start at the height above, with a vi = vf of the burn.
find the max height. at the top, vf=0
vf^2=0=vi^2+2ah where g=-9.8m/s^2, solve for h. Add that h to the heightattheendof burn