Finding max height for a rocket with upwards acceleration

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A toy rocket accelerates upward at 23 m/s² for 11 seconds before the motor stops, and the goal is to find its maximum height, disregarding air resistance. The initial calculations yielded a height of 1391.5 meters for the first phase of the ascent. To determine the additional height gained during free fall, the final velocity at the end of the first phase was calculated to be 253 m/s. Using this velocity as the starting point for the second phase, the calculations aimed to find the total height but resulted in confusion regarding the correct application of kinematic equations. The final answer should be expressed in kilometers, as specified in the problem statement.
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?
 
Did you answer in terms of km, or in terms of m? :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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