A model rocket is launched straight upward with an initial speed of 45.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 110 m. What can you say about the motion of the rocket after its engines stop? What is the maximum height reached by the rocket? How long after liftoff does the rocket reach its maximum height? How long is the rocket in the air?
rocket equation in first stage:
h = 1.5t^2 + 45t + c, but at launch, t = 0 and h = 0
so c = 0
h = 1.5t^2 + 45t
v = 3t + 45
when do the engines stop? When h = 110
1.5t^2 + 45t - 110 = 0
t^2 + 30t = 220/3
t^2 + 30t + 225 = 220/3 + 225 = 298.33333..
(t+15) = 17.27
t = appr 2.2723 seconds to reach the cut-off stage
after t = 2.2723 sec, v = 3(2.2723) + 45 = 51.817 m/s
we need a new equation:
a = -4.9
v = 51.817
h = 110
h = - 9.8t^2 + 51.817t + 110 , where t is the time after 2.2723 s
we want the vertex of this parabola
the t of the vertex is -51.817/-9.8 = 5.2874 s
so the total time until max height = 5.2874+2.2723 s or 7.56 seconds
You do the rest of it:....
you want the time until it hits the ground, that is
-9.8t^2 + 51/817t + 110 = 0
solve for t, accept only the reasonable answer.
remember this is the time from engine stop to landing on the ground,
so you will have to add the 2.2723 sec of the first stage.
Wheeww!, Better check all that arithmetic, should have written it out first