How is this equation obtained on this page So I have been researching on air resistance and projectile motion, yet I don't get the last part of paper where the writer has a graph with the X and Z axis with a set of equations that I cant find.
How come X=x/(v_0^2/g) and the same for Y=y/(v_0^2/g).
I have included a picture to show you the graph and the equations above.
I also have the sources as a link
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html
picture
 A: It's just the dimensionless variables.
It's common practice in computational studies to first remove the dimension by choosing suitable natural units of the system, so that the computer needs only to deal with quantities with no dimension.
Try to follow the derivation of the author, starting from the equations with dimensions at the beginning, and then choose $v_0^2/g$ as the unit of length, and $v_0/g$ as the unit of time to obtain the equation of motions in dimensionless form.
Edit: OK I just read your link in details. In fact your problem with resistance $-cv$ is solved analytically. The computer is used only in the last stage to plot the graph.
So you got
$$x=\frac{v_0v_t\cos\theta}{g}\left(1-e^{-gt/v_t}\right)$$
$$z=\frac{v_t}{g}\left(v_0\sin\theta+v_t\right)\left(1-e^{-gt/v_t}\right)-v_tt$$
where $v_t=mg/c$.
It is not a good idea to set values for all parameters $v_0$, $v_t$, $\theta$, and $g$. Instead you can measure length in units of $v_0^2/g$ and time in units of $v_0/g$. In other words, define the "dimensionless" lengths by
$$X=\frac{x}{v_0^2/g}$$
$$Z=\frac{z}{v_0^2/g}$$
and "dimensionless" time by
$$T=\frac{t}{v_0/g}$$
Then you have
$$X=\frac{v_t}{v_0}\cos\theta\left(1-e^{-(v_0/v_t)T}\right)$$
$$Z=\frac{v_t}{v_0}\left(\sin\theta+\frac{v_t}{v_0}\right)\left(1-e^{-(v_0/v_t)T}\right)-\frac{v_t}{v_0}T$$
Now you see you only need to set two parameters, viz., $v_0/v_t$ and $\theta$ when plotting the graph. Isn't that much simpler?
