physics
posted by ben on .
A model rocket is launched straight upward
with an initial speed of 45.4 m/s. It accelerates with a constant upward acceleration of
2.98 m/s
2
until its engines stop at an altitude
of 150 m.
What is the maximum height reached by
the rocket? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m

Vf^2 = Vo^2 + 2g*d,
Vf^2 = (45.4)^2 + 19.6*150 = 5001.16,
Vf = 70.7m/s = Vo for free fall phase.
h = ho + (Vf^2  Vo^2) / 2g,
h = 150 + (0  (70.7)^2) / 19.6=405m. 
When does the rocket reach maximum height?
Answer in units of s 
How long is the rocket in the air?
Answer in units of s 
Correction to your 10211,11:32am post:
The acceleration for the 1st phase should be 2.98m/s :
Vf^2 = Vo^2 + 2a*d,
Vf^2 = (45.4)^2 + 5.96*150 = 2955.16,
Vf = 54.36m/s. = Vo for freefall phase
h = ho + (Vf^2  Vo^2) / 2g,
h=150 + (0  (54.36)^2) / 19.6 = 301m.
t1 = (Vf  Vo) / a,
t1 = (54.36  45.4) / 2.98 = 3.0s. to
reach 150m.
t2 = (0  54.36) / 9.8 = 5.55s To go
from 150m to max. height(301m).
t(up) = t1 + t2 = 3 + 5.55 = 8.55s
d = V0*t + 0.5g*t^2 = 301m,
0 + 4.9t^2 = 301,
t^2 = 61.4,
t(dn) = 7.84s.
T = t(up) + t(dn) = 8.55 + 7.84 = 16.4s = Time in flight.