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i originally posted this question over at the math section but a guy told me to switch it to here instead.

i'm writing an assignment on firework rockets and their trajectory. Now of course im doing this with a lot of limitation as a realistic rocket calculation would be impossible to execute, at least for me.

I derived the classic rocket equation that assumes theres no gravity or aerodynamics through a simple differential equation. I then through simple calculations determined its trajectory if it was flying straight up.

Now what i want to do is derive a new equation that includes the drag formula, (Assuming the rocket flies straight up).

I'm a bit lost as to how I'm going to do this. The purpose it to find the difference between the normal rocket equation and one with drag integrated into it. I've searched all over for something like this but i haven't found anything.

Any ideas?

Rocket equation: $\Delta$$v$=$u*ln(\frac{M_0}{M})$.

Drag equation: $F_D=1/2*\rho*v^2*C_D*A$.

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  • $\begingroup$ Simply incorporate the force in the derivation step involving equation of change of momentum and impulse. $\endgroup$ – QuIcKmAtHs Dec 19 '18 at 2:52
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The classic derivation of an ideal rocket equation is as follows:

$\Delta mv=Mdu - vdm$

Net force acting on system: $(p-p_0)A-Mgcos\theta$, where $\theta$ is the angle of tilt of the rocket. Note that this step does not account for air drag. We shall simply add the air drag term into this equation, giving $(p-p_0)A-Mgcos\theta-F_D$.

We then equate change in momentum with force $dt$, giving us $Mdu=[(p-p_0)A-Mgcos\theta-F_D]dt + vdm=[(p-p_0)A-Mgcos\theta-F_D+\dot mv]dt$.

Using our knowledge of specific impulse, we can derive the equivalent exhaust velocity of a rocket given by $v_{eq}=\frac{(p-p_0)A-Mgcos\theta-F_D}{\dot m}+v$.

The follow steps will be as follows, such that we arrive at the equation $\Delta u=V_{eq}ln(\frac{M_0}{M})$, same as what you have gotten, only that $v_{eq}$ is different now.

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