# Max altitude of a model rocket

## 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: but i have looked around to see it it was right, and i found this equation: they give me different results, who is right, or are we both wrong?

Hope you can help

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What did you use as your initial differential equation for the acceleration of the rocket ##(\frac {d^2x} {dt^2})##?

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 so dx/dt, and its the same for the example i found

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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?

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 and do this 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?)

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.

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.

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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.

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

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