Ehhhhh.... this really isn't a "homework" problem but something derived out of curiosity(But I guess it could be considered homework-esque). 1. The problem statement, all variables and given/known data Basically, not accounting for drag or friction, what'd be the final velocity of a projectile launched from a compressed air chamber of volume v_0 connected DIRECTLY to a tube of radius R with launch length L with initial pressure P_0, with the projectile launcher set at angle [tex]\theta[/tex] from the horizontal? 2. Relevant equations m is the mass of the projectile g is the gravitational acceleration. x will be the length from the beginning of the tube to the given position of the projectile inside the tube. Area = pi*R^2 P_x = V_0*P_0/(V_0+x*Area) [The pressure where the projectile is down the tube at x-length] F = pressure*Area Work = [tex]\int[/tex] F ds (With the Y-axis perpendicular to the projectile's motion and X-axis parallel to it.) F_x = Area*P_x -mgsin([tex]\theta[/tex]) 3. The attempt at a solution The kinetic energy of the initial state of the projectile is 0. The net work done along the X-axis, that being along the length of the tube, will be the final kinetic energy. Since the net work is the integral of the Force applied along the X-axis(As there's no net Y_force with my set-up) in respect to distance, then K_f= Work = [tex]\int[/tex]_L,0 F ds = [tex]\int[/tex]_L,0 Area*P_x -mgsin([tex]\theta[/tex]) dx = Area*[tex]\int[/tex]_L,0 P_x dx -mgsin([tex]\theta[/tex])[tex]\int[/tex]_L,0 dx= Area*[tex]\int[/tex]_L,0 V_0*P_0/(V_0+x*Area) dx -L*m*g*sin([tex]\theta[/tex])= (Area is now A) V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex]) Now that the work ergo final Kinetic energy has been found, relating to the kinetic energy formula 1/2*m*v_f^2= K_f, v_f = (2K_f/m)^(1/2) = (2/m)^(1/2)*(V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex]))^(1/2) Is that right? I didn't think the solution would be that simple as I imagine a time-based solution would be much more difficult(even though I haven't even envisaged the solving procedures.). Some of the equations were derived from my own thinking, such as the pressure at length-x equation, so I fear some of my premises may be flawed and so the final solution.