Max altitude of a model rocket

  • #1

Homework Statement


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)

Homework Equations


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

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
 

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Answers and Replies

  • #2
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38
What did you use as your initial differential equation for the acceleration of the rocket ##(\frac {d^2x} {dt^2})##?
 
  • #3
What did you use as your initial differential equation for the acceleration of the rocket (##\frac {d^2x} {dt^2}##)? Or did you start with velocity (##\frac {dx} {dt}##)?
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
 

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  • #4
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38
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?
 
  • #5
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?
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?)
 
  • #6
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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.
 
  • #7
156
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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:
  • #8
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38
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.
 
  • #9
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.
Ok thank you. I will try that, and make an update post (if you have a little time later) :D
 
  • #10
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.
Sorry i said m(t)=-me*t but it's m(t)=m0-me*t
My mistake
 

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