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( Atc+Bc
 \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]