A simple function to draw an arc (I guess?) 
*

*sx = x starting point

*ex = x end point

*sy = y starting point

*ey = y max


having all the above, plus the current x, how do I calculate with a formula an y which isn't linear but that goes faster at first and slower approaching ey instead? (such as resembling gravity)
I guess I should use some sqr or some log somewhere, but I fail to see how exactly. I know this should be trivial but it looks like my basic math skills vanished, not having used them for too long, I'm sorry.
 A: There are many possible answers.  One cannot know what sort of answer is most suitable for your particular application without further information.  I will supply one answer that more or less fits your description.  In order to avoid confusion with multiplication, I will change your notation slightly, using for example $s_x$ instead of sx, $e_x$ instead of ex, and so on.
Here is a suggested formula.
$$y=e_y-(e_y-s_y)\left(\frac{e_x-x}{e_x-s_x}\right)^2$$
It is easy to check that this works.  When $x=s_x$, the squared term is equal to $1$, and $y=e_y-(e_y-s_y)=s_y$.  When $x=e_x$, the squared term is $0$, and $y=e_y$.  The fact that I squared ensures that the approach to $e_y$ is rapid at the beginning and slow when $y$ is close to $e_y$. I used the square function as a reaction to your mention of gravity. One can adjust the rate of approach by changing the exponent from $2$ to something else.  For a more rapid approach at the beginning, use, for example, $4$ instead of $2$. You can use non-integer exponents, like $1.6$ or $2.7$ instead of $2$ to adjust things so that they fit your application reasonably. 
