I would say engineering myself. I have watched a LOT of bottles thrown. I find it really tedious. This is how I think it works, and how I might model it.

Whether it will land or not depends on whether the motion of the base of the bottle is more or less vertical at the moment of landing.

Try to land a bottle by throwing it without flipping. You soon see that the problem is that a parabola has constant horizontal motion, so the bottle will always slide on landing.

The flip is essential so that the base of the bottle at the moment of landing is just before vertical, the mass of the water then forces is down — there is little or no horizontal movement at the moment of impact so the bottle doesn’t slide, and the rotational movement is still backwards if just before vertical so the bottle doesn’t tip over.

If the spin it too fast you get too small a margin for error at the moment of landing, so a slow spin is needed to maximise the sweet spot. One complete spin is easiest to control — the bottle starts vertical at the base of the throw, and ends vertical again.

Parametrically we have a classic parabola : x = t, y = 0.5 t^2 combined with a classic circular motion of a point rotating about the centre of x = sin t, y = cos t.

We can play around with variables modelling the spin speed inside the trig function, height of parabola etc, so a=Curve[t+sin(3 t),-0.5 (t-3)^2+4+cos(3 t) 0.5,t,0,6] in Geogebra shows the retrograde motion of the base relative to the centre of mass and how spinning too fast leaves little margin for error.

A solution around a=Curve[t+sin(0.6 t),-0.5 (t-3)^2+4+cos(0.6 t) 0.5,t,0,6] seems viable as a practical path, although in general the kids throw a lower loop.

Note if you spin the bottle the “other” way, replacing sin with -sin in the equation you get a curve that cannot cancel the motion without doing more than one gentle spin, which becomes impossible in practice.

I have taken no account of sloshing because 1) we are modelling at school level, not NASA level, and 2) because any throw with more than a little slosh tends to fail anyway, as the continued motion of the water in the bottle tends to make it topple even if you land it right.

I also haven’t taken into account the distance of the base of the bottle relative to centre of mass. Again, describing circular motion on top of parabolic motion is the aim: that is what little Maths there is in the this thing — the rest is engineering.

The amount of water is important because without enough water in it any plastic bottle will bounce and we need the mass of water pushing into the landing to keep it stead. But with too much the centre of mass is moved further from the base of the bottle, which makes the sweet spot for landing much smaller (as the mass of the water upon landing is much harder to cancel). But that’s physics, not Maths.

Feel free to delete if this comment is off topic for the thread Dan, but you asked how I might deal with it parametrically.