Runge-Kutta ODE Solver please keep in mind that the following is homework, I do not want answers, only help. If you are confused as to what I refer to in this question, or I don't make any sense, please refer to the link below. 
This is my assignment
I'm trying to plot the path of a firework that weighs $.0008 kg$ and has a density of $1900 kg/m^3$, and has a burn rate of $0.0027 kg/s$ in 2d using java. The firework has a launch velocity of $22 m/s$. The only force acting on the x-axis is the drag force, and wind velocity which I can change.
The forces acting on the x axis are:
-the initial velocity $22 m/s$
-the drag force
-gravity $9.807 m/s^2$
-the wind force: anywhere from $-15$ to $15$
The drag force is computed by $F_D=(pvAC_D)/2$
$p$=fluid density$=1.2 kg/m^3$
$v$=velocity magnitude 
$A$=cross sectional area$=0.00046140621$
$C_D$=drag coefficient$=.4$
Without any wind acting on the x axis, the velocity magnitude is merely: $\sqrt(22^2+0^2)=484$
Now using the formula: $F_D=(pvAC_D)/2$ we get: $F_D=(1.2*484*0.0004614*.4)/2=.05359$
Now we also add the mass times the gravity(9.8m/s^2)to get our total force on the y axis.
$F_y=-mg-F_D(v_y/v)$ ($v_y$, the velocity of y, is the same as v, the velocity magnitude, since the x axis velocity is 0, so $v_y/v=1$)
Which translates to $F_y=-0.0008*9.8-.5359*(1)=-.54374$
So there is a total of .54374 downwards force on the y axis. 
My professor then says to use the Runge-Kutta ODE solver to find the velocity of $x_k+1$ and $y_k+1$ but I'm not sure how to interpret the Runge-Kutta solver into the formulas that I've already been given. 
To conclude, I know the initial downward force on the object, I know that it has an initial velocity of 22 meters/s, and I know that the mass will be reducing as my path increases. I don't want to know the wind force at this time, as I want to add that in afterwards.
Could someone please help me understand how the Runge-Kutta ODE solver applies to this assignment? 
To answer Kalvotom:
The total force being placed on the sphere for each axis is as such:
$F_x=-F_D(v_x/v)$ and $F_y=-mg-F_D(v_y/v)$
quoting the assignment: 
"Since force equals mass times acceleration and acceleration is the first differential of velocity with time, these two equations can be expressed as two differential equations:"
$dv_x/dt=(F_Dv_x/mv)$ and $dv_y/dt=-g(F_Dv_y/mv)$
"These two ordinary differential equations cannot be solved using algebra, so we are going to have to resort to a numeric technique." AKA Runge-Kutta
 A: In order to see where the need for an ODE solver comes from, you have to be aware of the fact that you are trying to solve Newtonian equations of motion. I do not see those equations explicitly in your analysis. 
Edit: Hmm, let me try it this way (I do not want to solve your HW :-)): you want to solve
$$ z' = f(z), $$
where $z$ is a vector of unknown functions, let there be $n$ of them, and $f:\mathbb{R}^n\to\mathbb{R}^n$. The corresponding RK4 scheme (one of many RK methods available) works in the following way. Let $h$ be a small parameter (time step size) and let $z_0\in\mathbb{R}^n$ and $t\in\mathbb{R}$ be given. Then compute
$$ z_1 = z_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4), \quad t_{1} = t_0 + h, $$
where
$$ k_1 = h f(z_0), \ k_2 = hf(z_0 + \frac{1}{2}k_1), \ k_3 = f(z_0 + \frac{1}{2}k_2), \ z_4 = h f(z_0 + k_3). $$
This gives you a mapping $z_0 \mapsto z_1$. $z_0$ are the initial data at $t$ and $z_1$ is an approximation of the solution at time $t+h$. Now you just have to iterate this map over and over again. That is, you take $z_1$ instead of $z_0$ and compute $z_2$, etc.
In your case the unknowns are $z=(x,y,v_x,v_y)$. You have $x' = v_x$, $y' = v_y$, $v_x' = \cdots$, $v_y' = \cdots$. So if we denote $f=(f_1,f_2,f_3,f_4)$ then for example $f_1(x,y,v_x,v_y) = v_x$. You have to find the other componentes, write down this iterative scheme (essentially a simple loop where you record values at each iteration) and you are done.
