Inconsistency in units in differential equations for 2d projectile trajectory I am to solve the following two coupled second order differential equations involving the motion of a projectile. 

For the $y''(t)$ differential equation, I do not understand why the "$g/m$" term is not  just "$g$". The units do not match. I emailed my professor, and she responded that the term is supposed to be "$g/m$" and not "$g$", but she did not explain why. 
 A: For me, a valid line of reasoning is the following:


*

*The sum $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2$ only makes sense if $x$ and $y$ have the same units.

*Hence, the first equation yields $\frac{[x]}{\text{time}^2} = \frac{1}{\text{length}} \frac{[x]^2}{\text{time}^2}$, from which it follows that $[x] ( = [y]) = \text{length}$.

*Using the same reasoning as in 1., it follows that the dimension of every sum component on the right hand side of the second equation must have dimension $\frac{[y]}{\text{time}^2} = \frac{\text{length}}{\text{time}^2} = [g]$.


This suffices to conclude that $\frac{g}{m}$ should be $g$.
Note that the mistake could also have been made in the term $\frac{C_d A \rho}{2m}$. If we assume that the term $\frac{g}{m}$ is correct after all, this implies that the dimension of the constant multiplying $\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)^\frac{1}{2}$ should have dimension $\frac{\text{mass}}{\text{length}}$. For example, a constant like $\frac{C_d A \rho}{2m^2}$ does the trick. Then, it follows that $[x] = [y] = \frac{\text{length}}{\text{mass}}$, and the dimensions of the equation are correct.
