Towards the lower end
- Your rocket is slamming into a wall of air.
- Altitude has a quadratic response to reduction in velocity.
Part 1: Getting up to speed with the calculations you have probably already performed.
For instance, it would be nice to know what speed the rocket is moving at when the engine burns out. While gravity and drag also acts upon the rocket during these initial six seconds, the engine should be the most important part here.
The rocket equation comes in handy as usual, provided that we have an initial mass of 19.1kg and a dry mass of 10.6kg. We don't have the exhaust velocity, but that should be the same as thrust divided by mass flow, about 1800m/s.
All in all, I get a propulsive delta-v of 1060 m/s, which if we subtract the ~60m/s of gravitational slowdown over six seconds should be a roughly 1km/s top speed before we start considering drag. Oh, and the vertical component is reduced by some 3.4% by the launch not being straight up.
A parabolic arc should be sufficient to model this. We aren't getting high enough for the weakening of Earth's gravity to start playing any significant role.
I too get an altitude of 47.5km from that freefall arc, plus an additional 3km the rocket moved before the engine burned out, for a total of ~50km.
Part 2: a wall of air.
Conveniently, you have provided all the required parameters for the drag equation. This is by far the most imprecise part of it all, as supersonic fluid dynamics is not very "pen and paper" friendly. But sure, a simple drag model can help quantify what forces that are in play.
Contrary to your constant air density approach, I'm going to take a sneak peek at the instantaneous conditions at burnout first:
Firstly, at roughly 3km, the air density is more like 0.91kg/m³, increasing drag by 50% at the point in the flight were it is undoubtedly the most significant.
As we have all the numbers, the drag equation says the initial drag is about 2400N.
Which is nearly the same as the thrust the engine provides!
That means that we have to revisit the initial velocity from part 1, since drag during the initial six seconds is clearly eating a large part of it.
This is the point where differential equations are appropriate, but since this is "estimation", we can instead fiddle with the speeds a bit, and notice that around 700m/s the drag is eating up about half of engine thrust, and given the losses up to that point, that's about as far as its going to make it towards 1km/s.
Ballistically, that's already cutting apogee in half, still assuming the rest of the trajectory is drag free.
"drag free" is far from the case though, currently travelling at over mach 2 somewhere below 3km, drag outmatching gravity by a factor of about thirteen.
Even when the speed is eventually halved, drag is still over 3x stronger than gravity. Meaning that the partition of the flight that should cover 3/4 of the apogee altitude gain lasts for at least 4x less time that it would without drag.
By these observations alone, the upper bound of the apogee is already getting pretty close to your second estimate.
