# Max altitude of a model rocket

• Januz Johansen
In summary, the task was to determine the maximum height of a model rocket launched at a 45 degree angle with a finale velocity of 15.69 m/s and an extutevelocety of 14.23 m/s. The total mass of the rocket is 180 grams, with an empty mass of 52 grams and a mass per second of 0.45 kg/s. The total burn time is 0.28 seconds. The attempt at a solution involved integrating the rocket equation to get s(t) and then using Pythagoras' theorem, but this method yielded different results compared to using the equation found in an example. It was later suggested to replace every t in the equation with the function for m(t) and
Januz Johansen

## 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})##?

TJGilb said:
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?

TJGilb said:
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.

Last edited:
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.

TJGilb said:
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

TJGilb said:
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

## 1. What factors determine the maximum altitude of a model rocket?

The maximum altitude of a model rocket is determined by several factors, including the size and weight of the rocket, the amount of thrust produced by the engine, and the wind conditions at launch. Other factors such as the angle of launch, the quality of the rocket's construction, and the type of engine used can also affect the maximum altitude.

## 2. How do you calculate the maximum altitude of a model rocket?

The maximum altitude of a model rocket can be calculated using a mathematical formula that takes into account the rocket's mass, thrust, and drag. This formula, known as the Tsiolkovsky rocket equation, is often used by scientists and engineers to predict the performance of rockets.

## 3. What is the typical maximum altitude of a model rocket?

The maximum altitude of a model rocket can vary greatly depending on the factors mentioned above. However, on average, a model rocket can reach a maximum altitude of 100 to 1500 feet. Some high-powered rockets can reach altitudes of up to 10,000 feet or more.

## 4. How do you ensure the safety of a model rocket at maximum altitude?

To ensure the safety of a model rocket at maximum altitude, it is important to carefully follow the instructions provided by the rocket's manufacturer. This includes using the correct engine size, launching the rocket in an open and clear area, and avoiding launching in windy conditions. It is also important to always have adult supervision when launching a model rocket.

## 5. Can a model rocket reach outer space?

No, a model rocket is not capable of reaching outer space. The minimum altitude required to be considered in outer space is 62 miles, which is much higher than the maximum altitude that a model rocket can achieve. Only specialized rockets with advanced technology and powerful engines are able to reach outer space.

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