The Attempt at a SolutionLaunching 1000kg Weather Rocket: 5100m in 20s

[hachi_roku]

Homework Statement

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

Homework Equations

The Attempt at a Solution

i don't know what formula to use. I am trying to set up systems where final position is 5100m, and initial position is 0m. please help

i am trying to go about this saying that initial velocity was zero and using the equation
final velocity = initial velocity + (a)(t) where a is acceleration and t is time.

 

[Stovebolt]

You have two distinct phases for the motion of the rocket (with the motor accelerating the rocket and without). The rocket will be changing position in both of these cases, but at different rates. What you do know, is that the sum of the position changes in each phase will be equal to 5100 m.

Try setting up two separate equations, then solve one in terms of the other. It will help you if you draw free body diagrams of the forces acting on the rocket for each phase.

Give it a try and post what you come up with.

 

[Hunterbender]

What are you trying to solve?

 

[Delphi51]

Amazing, you can figure out everything with that little bit of information!
In addition to Stovebolt's suggestion, I would write a d = AND a v = formula for each phase of the motion.

Careful, don't assume the rocket is stopped at time 20 s.

 

[hachi_roku]

edit: forgot to say what I am trying to figure out hehe

what is the acceleration during the first 16s?