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A model rocket

  1. Jan 22, 2015 #1
    Hello loves, I know there are post very very similar to this, but I just needed some help understanding it. any help is truly appreciated.

    1. The problem statement, all variables and given/known data

    A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 30 m/s2 until its engines stop at an altitude of 150 m. How long is the rocket in the air?

    Vi= 50m/s
    Vf= ??
    a= 30 m/s^2
    Change in y= 150

    2. Relevant equations
    upload_2015-1-22_23-33-37.png

    3. The attempt at a solution
    Alright so, I think to solve this problem, its a 2 step thing, this is caused because of the 2 accelerations, one from the rocket's fuel, and then once it empties, the second one from earth's gravitational acceleration. I know I have 3 givens so I can find the other 2 variables. Alright this has led me to choosing the fourth and first equation. Although im not sure if i can replace the delta x with delta Y, because the rocket would be going up into the air vertically, so it would be y instead of x right? this is where i am stuck, i dont really know what to do once i get to this point.

    a preemptive thank you guys for the help
     
  2. jcsd
  3. Jan 23, 2015 #2

    gneill

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    The particular symbols used to represent quantities aren't as important as the equation matching the given situation. Here you have motion in a straight line with constant acceleration(s). The equations you've shown match the scenario, so feel free to change Δx to Δy, or y, or z for that matter.
     
  4. Jan 23, 2015 #3
    Alright, so, the Δx is replaceable, that is a good thing to know. Alright so I have been manipulating all of these equations to try to isolate time. So for the first part, I am trying to find out how long it takes to get to 150 meters. I have yet to find this equation, or been able to manipulate these 4 to give it to me. for the second part, I can find the Final velocity of the first part, and make that the initial velocity, and the final velocity of the second one will be 0 correct? because it would hit the floor coming to a stop. So once again i have to isolate t, but this time i have the following variables:
    Yi = 150 meters
    Yf= 0
    a=-9.8 m/s^2
    Vi= Vf of part 1
     
  5. Jan 23, 2015 #4

    haruspex

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    In a constant acceleration problem, there are typically five variables: initial speed, final speed, acceleration, distance and time. Given any three you can calculate the remaining two. Accordingly, there are five SUVAT equations, each involving four variables.
    For the 150m stage, which three variables are known? Pick the equation that involves those three and time.
    But you will anyway be needing to calculate the speed at that altitude too.
     
  6. Jan 23, 2015 #5
    Okay, so i used the second equation to to solve for t, using the quadratic formula. so i got that t=1.90793s. I then used the fourth equation to solve for Vf which came out to be Vf=107.238 m/s. So now I am left only with 2 variables, Vf which now becomes Vi since i am now at the second phase where the engine stops. The second one i have is -g which is -9.8 m/s^2. I know I am overlooking something but I just cant figure it out. Unless Δy is the other variable, but I only have it as initial at 150 for phase 2, and I am not told it started from the ground, infact i am kind of told its not, because it is already going at a velocity of 50m/s. So i am not sure what to do
     
  7. Jan 23, 2015 #6

    haruspex

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    It's now at 150m. What will y be when it lands?
     
  8. Jan 23, 2015 #7
    Well it would be 0, because it lands right? So -150 would be Δy? this wouldnt make sense though because it would continue to go up for a little bit before it would start to go back down, right?
     
  9. Jan 23, 2015 #8

    haruspex

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    That doesn't matter, the equations will take care of it. The acceleration is constant from the time the engines stop until it hits the ground. Note that you will get a quadratic equation for that time, but only one solution will make sense.
     
  10. Mar 19, 2015 #9
    Oh, I forgot about this, I was able to get it, thanks guys
     
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