find the speed at engine shutdown, and the height of it at that time.
H=157 m given
vf^2=2ah solve for vf, a, h given.
knowing vf at shutdown, what is the height it now goes?
speed at the top is zero, so
0=Vfabove^2 + 2ah where a is now -9.8m/s^2, solve for h
final height: h+157
A model rocket is launched straight upward with an initial speed of 47.7 m/s. It accelerates with a constant upward acceleration of 2.02 m/s2 until its engines stop at an altitude of 157 m. What is the maximum height reached by the rocket?
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