usahockey7 said:
And it's difficult to work with, without using a program?
Yes. If there is an easy way to solve the DEs analytically, I'm not aware of it.
usahockey7 said:
The easiest way to solve it I guess is to use a program but I have no knowledge of programming, I've never taken a programming course throughout high school.
Right. Are there any languages you are familiar with? I prefer MATLAB for doing numerical analysis, but I doubt you'll have access to it. There is a free clone called Octave that is similar to MATLAB. Other good candidates are NumPy, a numerical version of Python (I'm only tangentially familiar with it) and Sage.
usahockey7 said:
I think he wants us to use the drag equation, but I've heard it's not a linear function?
No, the drag equation is a quadratic function. When you solve the equations of motion you get these second order nonlinear ordinary differential equations (bear with me for a moment):
<br />
\begin{align*}<br />
\ddot{x}&= -\frac{b}{m}vv_x \\<br />
\ddot{y}& = -g - \frac{b}{m}vv_y<br />
\end{align*}<br />
The double dots on x and y mean the second derivative of x and y, which is acceleration. A derivative, if you are unfamiliar with them (I'm not sure what's covered in your pre-cal class, some will introduce derivatives at the end), gives the rate of change of a function. It is essentially the "slope" of a function.
Now, the problem here is that to solve these equations numerically we need them in the form y' = f(t,y). This means we will need to solve four equations simultaneously. First, we find velocity:
<br />
\begin{align*}<br />
\dot{v} _x&= -\frac{b}{m}vv_x \\<br />
\dot{v} _y&= -g - \frac{b}{m}vv_y<br />
\end{align*}<br />
Then we use the velocity to find the position:
<br />
\begin{align*}<br />
\dot{x} &= v_x(t) \\<br />
\dot{y} &= v_y(t) \\<br />
\end{align*}<br />
Then we use a solver to find the solutions to these equations. We'll need initial conditions, which we'll get from your work on the earlier part of the problem.