Hello all, I recently made some model rockets and thought it would be fun to predict their height before setting them off this weekend. I thought it would also be interesting to compare the difference in the predictions of a model with no drag and a model with quadratic drag. Surprisingly, the model without drag predicted a lower height. I probably made some error somewhere, but I don't see it and I was wondering if anyone could offer some insight. I'll provide all the information and hope there isn't some simple answer that I've overlooked. Thanks in advance for anybody even willing to look at this rather lengthy post. The simplifying assumptions I've made for both models during the boost phase are: 1) Use the average thrust of the engine 2) Use the average mass of the rocket+engine+fuel The thrust of the engine is 4.2N The average mass of the rocket is 0.08kg The burn time of the engine is 1.1s No Drag This is a simple exercise in kinematics. The equation of motion is T-mg=ma where T is the average thrust and m is the average mass. Solving the differential equation predicts a velocity of 46m/s and an altitude of 26m when the engine has used up its fuel. The equation of motion during the coasting phase is ma=-mg. Solving this for the total height gives 136m. Quadratic Drag My drag coefficient (1/2 rho * Cd *Area) is equal to 0.00035 kg/m The equation of motion is ma=T-mg-cv^2. To simplify the integration I solved for the terminal velocity using this equation which is about 97m/s. The solution to the integral solving for the velocity of the rocket at burnout is Subscript[v, b]=Subscript[v, ter]((1-E^-k\[Tau])/(1+E^-k\[Tau])) where tau is the burn time of the engine and k is a constant that cleaned up the mess of the variables a bit. The height when the engine burns out was solved similarly and came out as Subscript[y, b]= \!\(\*OverscriptBox["m", "_"]\)/(2c) ln (\! \*SubsuperscriptBox[\(v\), \(ter\), \(2\)]/(\!\( \*SubsuperscriptBox[\(v\), \(ter\), \(2\)] - \*SubsuperscriptBox[\(v\), \(b\), \(2\)]\))) (hope this is readable, I'm cutting and pasting from another document). Anyway the predictions are close to that of the burnout phase for no air resistance. This model gives 43m/s and 25m for the velocity and height. For the coast phase the equation of motion is ma=-mg-cv^2. Once again I solved for the terminal velocity to simplify the equations (different than the vter above) Going through the process to find the height gives the expression Subscript[m, c]/(2 c) ln(-\! \*SubsuperscriptBox[\(v\), \(ter\), \(2\)]/(\!\( \*SubsuperscriptBox[\(v\), \(b\), \(2\)] + \*SubsuperscriptBox[\(v\), \(ter\), \(2\)]\))) where m is now the mass of the rocket without fuel, c is my drag coefficient, vter is now equal to about 46m/s and vb is the velocity at burnout. Anyway, the total height predicted is now close to 270m.