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Max altitude of a model rocket

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello all, I have to determine the maximum height of a model rocket.
    The rocket is launched at an 45 degree angle
    The finale velocity is 15,69 m/s (v)
    the extutevelocety is 14,23 m/s (c)
    The total mass is 180 gram (m) or m0)
    The emty mass is 52 grams (mf)
    The mass pr sek is 0,45 kg/s (me)
    And the total burntime is 0,28 sek (t)

    2. Relevant equations
    How can i calculate the hight at burnout, (i know how to calculate the last part) or all in one?

    3. The attempt at a solution
    I have tryed intregrating the rocket equation (v(t)) to get s(t), so i can calulate it with phytagoras, i got the equation from v(t) to:
    upload_2016-12-9_17-49-31.png
    but i have looked around to see it it was right, and i found this equation:
    upload_2016-12-9_17-58-14.png
    they give me different results, who is right, or are we both wrong?

    Hope you can help
     

    Attached Files:

  2. jcsd
  3. Dec 9, 2016 #2
    What did you use as your initial differential equation for the acceleration of the rocket ##(\frac {d^2x} {dt^2})##?
     
  4. Dec 9, 2016 #3
    Hello, Yes sorry forgot, i used the velocity upload_2016-12-9_18-42-29.png so dx/dt, and its the same for the example i found
     

    Attached Files:

  5. Dec 9, 2016 #4
    So you know velocity is the derivative of position with respect to time, but in your solution it looks like you integrated with respect to mass (or maybe turned time into a function of mass), am I correct?
     
  6. Dec 9, 2016 #5
    I see in the examble i found he rewrite the equation to
    upload_2016-12-9_18-59-26.png
    and do this
    upload_2016-12-9_19-0-6.png
    but will this make the difference, it should be the same (m(t) =mf at the burnout time, because there are no more fuel right?)
     
  7. Dec 9, 2016 #6
    When deriving the rocket equation, you start with the sum of forces to get ##\sum F =thrust-mg=ma## which can be rewritten as ##\frac {d^2x} {dt^2} = \frac {thrust-mg} m##. But since mass is a function of time in rockets, it can be replaced by ##\frac {d^2x} {dt^2} = \frac {thrust-{m(t)}g} {m(t)}##. For the record thrust is the mass flow rate multiplied by the exhaust velocity.

    It actually looks like they simplify it a little by not including the initial mass in their equation for m(t). But if you use the correct mass equation of ##m(t)=m_0-mfr*t## then solve for time you'd get ##t= \frac {m(t)-m_0} {mfr}## where the mfr is the mass flow rate. If you replaced every value of t with that "stuff" and integrated with respect to it, that could work. But that would be really overcomplicating the integration.
     
  8. Dec 9, 2016 #7
    Or going off what they use for m(t) you could integrate with to ##\frac {m(t)} {-m_e}## after plugging that into every t value.
     
    Last edited: Dec 9, 2016
  9. Dec 9, 2016 #8
    I was actually looking at what you got a bit closer, and it might be good. I'd recommend replacing every t in their equation with your function for m(t), simplifying, and see if it matches what you got.
     
  10. Dec 9, 2016 #9
    Ok thank you. I will try that, and make an update post (if you have a little time later) :D
     
  11. Dec 9, 2016 #10
    Sorry i said m(t)=-me*t but it's m(t)=m0-me*t
    My mistake
     
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