First I want to know the path the shell will follow through space. It's like putting a bright light on the shell, and photographing the entire flight with the camera's shutter stuck open. When I develop the film, I see a portion of a parabola that begins at the cannon and ends where the shell hits the ground.
I start with the basic laws of motion in the x and y direction. I will assume that the projectile starts at coordinate x=0, y=0, and ignore air friction. If "v0" is the initial velocity, and "angrad" is the angle of inclination of the velocity vector (cannon barrel) from the horizontal, then the x location at any time t after firing is:
x = v0* cos(angrad) * t (1)and the y location at the same time is:
y = v0 * sin(angrad) * t - 0.5 * g * t**2 (2)for our problem v0=300 meters per second (about sound speed) and g =9.807 m/s**2
I solve Equation (1) for t and get:
t = x / (v0*cos(angrad))
Now I substitute this into Equation (2) and get:
y(x) = x*tan(angrad) - (0.5*g*x**2)/(v0*cos(angrad))**2
This is the equation for the light streak in our picture. and we want to locate the point other than x=0 for which y(x)=0. Setting y(x)=0 in the above equation and factoring gives:
x*( tan(angrad) - 0.5*g*x/(v0*cos(angrad))**2) = 0
The solution of interest is when:
0.5*g*x/(v0*cos(angrad))**2 = tan(angrad)
x = 2*v0**2*sin(angrad)*cos(angrad)/g
Since trigonometric functions are expensive, I'll substitute my knowledge of trigonometric double angle formulas into the above equation to get the final expression for the value of x at which the projectile crashes back into the ground.
x = (v0**2)*sin(2*angrad)/g
Yes there is an easier way to get this answer, but this approach will be helpful soon when we look at ground that isn't flat.