A bullet is fired from a gun: what angle of elevation of the gun will result in the bullet traveling the maximum distance before hitting the ground? If you ignore the effects of air resistance, the bullet travels in a parabolic arc and it is easy to compute that the optimum firing angle is 45 degrees. However, if you consider the effects of air resistance, this becomes a much more difficult problem -- one for which there is no simple analytical solution.

This example uses NLREG to compute the firing angle that optimizes the horizontal distance traveled, taking into consideration the effects of air resistance. Since there is no analytical function to compute the distance as a function of the angle, the NLREG procedure uses an iterative procedure to simulate the flight of the bullet and determine the horizontal distance traveled for a specified angle. This procedure uses the "finite differences" method to compute the effects of gravitational acceleration and air resistance forces on the bullet over short steps of time. For each angle, the procedure iterates until the bullet hits the ground and then determines the horizontal distance traveled at the point of impact.

If you study the example, you will see that the force due to air resistance is proportional to the square of the speed. In this example, the initial muzzle velocity was chosen to be 300 meters/second. The air resistance coefficient was selected so that the bullet would have a terminal velocity (if dropped from a very high height through the air) of about 90 meters/second. If you set the air resistance coefficient to 0 (which has the effect of ignoring air resistance) you will find that the computed optimum angle is about 45 degrees. But with air resistance, the optimum angle is about 32 degrees.

   Title "Firing angle for maximum range";
    /* Parameter whose optimum value is to be determined */
    Parameter Angle=40;        /* Firing angle */
    Constrain Angle=20,70;     /* Put reasonable range on angle */
    /* Constants for problem */
    Constant Vm = 300;         /* Muzzle velocity (m/s) */
    Constant G = 9.80;         /* Acceleration of gravity (m/s^2) */
    Constant Airres = 0.0012;  /* Air resistance factor */
    Constant Step = 0.005;     /* Time step (seconds) */
    /* Work variables */
    Double Vx,Vy,X,Y,V,Dar;
    Double Xlast,Ylast;
/*
 *  Projectile is fired from ground level.
 */
    X = 0;
    Y = 0;
/*
 *  Compute initial X,Y velocities.
 */
    Vx = Vm * cosd(Angle);
    Vy = Vm * sind(Angle);
/*
 *  Simulate the flight of the projectile by advancing time
 *  through small steps until it hits the ground.
 */
    do {
/*
 *  Compute new X,Y position using current velocity and
 *  time step.
 */
        X += Step * Vx;
        Y += Step * Vy;
/*
 *  Apply gravitational acceleration to Y velocity.
 */
        Vy -= Step * G;
/*
 *  Apply air resistance to speeds.
 *  Note: Force of air resistance is proportional to the
 *  speed squared.
 */
        /* Compute total linear speed */
        V = sqrt(Vx^2 + Vy^2);
        /* Determine deceleration due to air resistance */
        Dar = Airres * V^2;
        /* Apply this to the X and Y velocities */
        Vx -= Step * Dar * Vx / V;
        Vy -= Step * Dar * Vy / V;
/*
 *  Remember last position above the ground.
 */
        if (Y > 0) {
            Xlast = X;
            Ylast = Y;
        }
/*
 *  Loop until projectile hits the ground.
 */
    } while (Y > 0);
/*
 *  Projectile hit the ground (and may be underground).
 *  Use last above-ground position to interpolate X
 *  at the point of impact (i.e., when Y was 0).
 *  Assume it moved in a straight line during the last step.
 */
    X = Xlast - Ylast * ((X - Xlast) / (Y - Ylast));
/*
 *  Maximize the distance.
 *  Note: The absolute value of the function is minimized
 *  which results in X being maximized.
 */
    Function 100000 - X;
    Data;