@mes's thread
Problems Plus 5: Launch Angle of 56° maximizes TOTAL distance a projectile travels
In this video, I show that firing a projectile has a maximum total distance traveled in the air when the launch angle is approximately 56°. This is 11° higher than the 45° angle needed to maximize the total horizontal distance. I derive this by starting from the parametric equations of trajectory, obtaining the velocity vector as the derivative of the position vector, and then obtaining the integral formula for the arc length the projectile travels (whose integrand is the magnitude of the velocity vector). The arc length is maximized when its derivative is zero, i.e. at a critical point, thus obtaining our answer of about 56° and a max total distance of about 1.20v²/g. Fascinating stuff!
