Finding max height for a rocket with upwards acceleration

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garcia1
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Homework Statement


A toy rocket, launched from the ground, rises
vertically with an acceleration of 23 m/s2 for
11 s until its motor stops.
Disregarding any air resistance, what max-
imum height above the ground will the rocket
achieve? The acceleration of gravity is
9.8 m/s2 .
Answer in units of km.


Homework Equations



I used kinematics equation: x=Vo*t + 1/2at^2



The Attempt at a Solution



I was a little unsure how to go about this problem, since most free fall problems I've dealt with use only gravity as acceleration.

I tried plugging the following values into my equation:
Vo = 0m/s
a = 23 m/s^2
t = 11s

I got the answer 1391.5m, but this was wrong. I think the problem is simpler than I'm thinking about it, but I can really use some help on this.
 
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hi garcia1! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
garcia1 said:
I tried plugging the following values into my equation:
Vo = 0m/s
a = 23 m/s^2
t = 11s

I got the answer 1391.5m …

that's the height at the end of the first stage …

you still need the extra height that it goes with a = -9.81 m/s2 :wink:
 
So what I did next is solve for the final velocity of part 1 by using the fact that Vo = 0m/s since the rocket starts from rest. Using the equation V = Vo + at, I got the equation

V = 23*11 = 253m/s

With this I placed this final velocity as the initial velocity for the next segment. I determined in this 2nd segment that X = ?, Vo = 253m/s, a = -9.81m/s^2, and Vf = 0, since the rocket must come to rest at the final height before falling.

I got the following equation:
Vf^2 = Vo^2 + 2ax -> x = Vf^2 - Vo^2 / 2a.

This answer was 3265.76m. Adding this to the initial 1391.5m, I got 4657.27m. This was wrong though, so I think there is something in this second step I'm getting wrong. Any thoughts?