Dear Dr. Hwang,

                        I am Daniel Baldacchino, a second year mechanical engineering student at the University of Malta. I have an assignment to build a simple catapult to launch a 100g water balloon, as far as possible but also to try and predict as accurately as possible the range it will achieve with kinetic equations. Since this has to be an accurate calculation, I don't think we can afford to simply neglect drag on the balloon.

I now refer to your equations posted on the "Projectile Motion with Air Drag" forum.
                                            Fx= - b* V2 * (vx/|V|) = - b * vx * |V|;
                                            Fy= - b* V2 * (vy/|V|) = - b *vy * |V|;

What exactly are the terms (vx/|V|) ? initially i thought these were a position vector , but the fact you use vx has confused me.



Also,  when you plug this force into the acceleration equations ,

                                            dvx/dt = - b * ( Math.sqrt(vx*vx+vy*vy) / m ) * vx;
                                            dvy/dt = - b * ( Math.sqrt(vx*vx+vy*vy) * vy / m-g ;

how would you integrate to solve for these values? are vx and vy intial values or would you have to integrate them from some intial value to 0 ?

I was considering, for the sake of simplifying the calculation, to eliminate effect of drag in the vertical direction, and consider only a force in the horizontal direction, say Fx = -bv      ( or v^2, im not sure which to use, ive found both versions!)

therefore d(vx)/dt = bv/m

and using (ax)dx =  v(dv), i would integrate x from 0 to x, and v from some initial velocity to a final value. 
I would obtain another equation in the y direction, assuming a constant acceleration -g and try and find a way to solve...

Any insight into what your thoughts are would be much appreciated.
Thank you for your time :)

Best Regards,
Daniel Baldacchino