Equation of motion - choosing directions for the coordinate axis I'm not sure this equation is the right one if we invert the direction of the positive $y$-axis.
Am I right if we assume the positive direction of the $y$-axis downwards, that the correct equation of motion would be 
$$m\ddot{y}=mg-{\frac{\rho A C}{2}\dot{y}^2}$$
as the spacecraft tries to land downwards?
Where would be the direction of acceleration due to the net force?

 A: It is up to you to decide which coordinate system you'd like to use.
In your case, you have chosen the downward direction of the y-axis to be the positive direction. Therefore, the sign of the $mg$ will be positive.
Since the $mg$ force is acting downwards, the term will be positive. The force which is acting upwards will be negative.
$m\ddot{y}=mg-{\frac{\rho A C}{2}\dot{y}^2}$
Your diagram however, hints that you should use the upward direction as the positive direction. In this case, the signs of the quantities will flip.
$m\ddot{y}={\frac{\rho A C}{2}\dot{y}^2} - mg$
A: I believe that the original equation as presented in the OP is ill-formed.
A correct vector equation should not be affected by an arbitrary choice as to the positive direction for the vectors.
In the equation given, the LHS will change sign under such a change in the positive direction. A landing space ship that is slowing to a stop will have a numerically positive upward acceleration, or a numerically negative downward acceleration.
Likewise the acceleration of gravity (second term, RHS) will numerically change in sign.  If up is positive, $g=-9.8$: if down is positive, $g=9.8$
However, the drag term, with the square of the velocity, is always non-negative.  It does not switch signs, and the equation fails.
The inclusion of a unit vector in the drag term, with a direction opposed to the velocity, would make the equation correct...  
A: The correct form of the equation is
$$ m\ddot{y}=mg-{\frac{\rho A C}{2}|\dot{y}|\,\dot{y}} $$
or
$$ \begin{cases} 
  m\ddot{y}=mg-{\frac{\rho A C}{2}\dot{y}^2} & \dot{y}>0 \\
  m\ddot{y}=mg+{\frac{\rho A C}{2}\dot{y}^2} & \dot{y}<0 \\
\end{cases} $$
This is because air resistance must oppose the motion $\dot{y}$.
