Begin with a free-body diagram. The velocity has been included on the free-body diagram for reference.

To apply Eqs. 5, the x and y components of the total force on the projectile must be determined. Since v/|v| is a unit vector, D can be written as

The external forces are weight and drag,

The x and y components of the total force are

Now, consider the case when C = 0.002, and let Δt = 0.1. At the initial time t_o

     t_o = 0

     v_x(t_o) = 400 ft/s           v_y(t_o) = 400 ft/s

Using the four basic equations for position and velocity (Eqs. 5) the following is determined,

x - position
     x(t_o + Δt) = x(t_o) + v_x(t_o) Δt
     x(0 + 0.1) = 0 + 400 (0.1)
                    = 40 ft

y - position
     y(t_o + Δt) = y(t_o) + v_y(t_o) Δt
     y(0 + 0.1) = 0 + 400(0.1)
                    = 40 ft

                              x - velocity
                                  v_x(t_o + Δt) = v_x(t_o) + (1/m) ΣF_x(t_o, x(t_o), y(t_o), v_x(t_o), v_y(t_o))Δt
                                  v_x(0 + 0.1) = v_x(0) - (C/m) v_x(0) (v_x 0² + v_y 0²)^0.5 Δt
                                  v_x(0.1) = 400 - (0.002/100) (400) (400² + 400²)^0.5 (0.1)
                                             = 399.6 ft/s

                               x - velocity
                                   v_y(t_o + Δt) = v_y(t_o) + (1/m) ΣF_y(t_o, x(t_o), y(t_o), v_x(t_o), v_y(t_o))Δt
                                   v_y(0 + 0.1) = v_y(0) - g Δt - (C/m) v_x (0) (v_x 0² + v_y 0²)^0.5 Δt
                                   v_y(0 + 0.1) =  400 - 32.2(0.1) - (0.002/100) (400) (400² + 400²)^0.5 (0.1)
                                                     = 396.3 ft/s

Results Plot

Continuing this way, the results for the first five time-steps can be obtained as shown in the table at the left.

These values can then be plotted to obtain an approximate solution for the trajectory of the projectile:

When there is no drag (C = 0), the closed-form solution for the trajectory can be obtained and compared with the numerical solution.

The graph below presents this comparison using Δt = 3.5s, Δt = 1.0s, and Δt = 0.1s.

Notice that the numerical solution with Δt = 0.1s is the closest to the exact solution. This indicates that as the time-step is decreased, the numerical solution becomes more accurate.