posted by Haroula on .
A model rocket is launched straight upward with an initial speed of 58.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 120 m.
(a) What can you say about the motion of the rocket after its engines stop?
(b) What is the maximum height reached by the rocket?
(c) How long after liftoff does the rocket reach its maximum height?
(d) How long is the rocket in the air?
(a) At rocket cutoff, it changes from accelerating at 1.5 m/s^2 to decelerating at rate -g = -9.8 m/s^2.
(b) At 120 m cutoff, the rocket's velocity V is given by
V^2 - 58^2 = 2 a X = 360 m^2/s^2
V = 61.02 m/s
Maximum height occurs t = 61.02/9.8 = 6.23 s later. Additional height above 120 m is
(V/2)*t = 190.1 m higher, or 310.1 m
(c) already answered. See t.
(d) During acceleration, the time the rocket was burning was
120 m/(Vaverage) = 2*120/(58 +61)
= 2.02 s.
Add that to 6.23 s for the time to reach maximum height. Then add the time it takes to fall from there.