The solution of the 2D drag differential equitation is:
Horizontal component
[EQ]{\frac {d^{2}}{d{t}^{2}}}x \left( t \right) =-c \left( {\frac {d}{dt
}}x \left( t \right) \right) ^{2}[/EQ]
[EQ]x \left( t \right) ={\frac{1}{c} \ln \left( {\it \A}\,tc+{\it \B}\,c
\right)}[/EQ]
Vertical component
[EQ]{\frac {d^{2}}{d{t}^{2}}}y \left( t \right) =-g-c \left( {\frac {d}{dt
}}y \left( t \right) \right) ^{2}[/EQ]
The solution is
[EQ]y \left( t \right) = \frac {1}{2c}\,\ln \left( {\frac {c}{g} \left( E\cos \left(
\sqrt {{\it gc}}t \right) -F\sin \left( \sqrt {{\it gc}}t \right)
\right)} \right)[/EQ]
In all equitations, c is the short var for [EQ]c={\frac {\rho\,AC_{{F}}}{2m}}[/EQ]