physics
posted by desperate help on .
A model rocket is launched straight upward with an initial speed of 30.0 m/s. It accelerates with a constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 170 m.
(a) What is the maximum height reached by the rocket?
m
(b) How long after liftoff does the rocket reach its maximum height?
s
(c) How long is the rocket in the air?
s

Maximum height occurs where the velocity is temporarily zero.
While accelerating, the height is
y(t) = 30 t + 1.25 t^2
and the speed is
v(t) = 30 + 2.5 t
First figure out when y = 170m, and what the speed is at that time.
1.25 t^2 +30t 170 = 0
t^2 + 24t 136 = 0
That does not factor easily, so use the quadratic formula.
t = (1/2)(24 + 33.47) = 4.73 s
v(t=4.73) = 30 + 11.83 = 41.83 m/s
Zero velocity will be reached after an additional time t', such that
g*t' = 41.83 m/s
t' = 4.27 s.
The total time after liftoff is then 9.00 seconds
The maximum altitude reached is
170 m + 41.83 t' (g/2)t'^2
= 170 + 178.6  89.3 = 259.3 m
(b) t + t' = 9.00 s
(c) 9.00 s + (time to fall from 259.3 m)