Ballistics is not easy.

[honkytonk]

I return to this code, the other formula justbasiccom.proboards.com/thread/891/parabolic-rescue?page=1&scrollTo=6249 is not good. (the projectile falls further for 70° than for 45°, with constant v0).
I think the horizontal line after landing is the rest of the energy.
because it is shorter for 85° than for 45°.
In this line:
"fy =0 -g*m -0.5*D*vy^2*Cd *A '*sgn(vy)"
we see "m", which would be the initial velocity calculation, but there is no energy, it's weird.
We'll have to think to understand this knot bag.

[honkytonk]

The trajectory of a projectile is parabolic, but is not a parabola.
Because: The parabola is symmetric, while the trajectory is not:
At the top of the curve, the projectile has consumed all of its energy (since it is falling).
In the descent, the fall is necessarily faster than the horizontal displacement.
The code curve of this thread is consistent.
It remains to peel it to integrate my parameters.