Problems Plus 5: Launch Angle of 56° maximizes TOTAL distance a projectile travels

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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!

Timestamps

Problem 5: Total distance traveled by projectile – 0:00
Solution: Obtain position vector, velocity vector, and magnitude of velocity vector – 1:31
Completing the square to simplify the magnitude of the velocity vector – 8:19
Projectile hits the ground when y = 0 – 14:06
Distance traveled is the arc length integral formula – 15:36
Solving integral using Formula 21 of the Table of Integrals and a LOT of algebra! – 17:57
Arc length is maximized when its derivative is zero, i.e. it's a critical point – 40:00
Solving for the angle using Grok AI to get α ≈ 0.9855 radians ≈ 56°: https://grok.com/share/c2hhcmQtMg_e5105c99-e220-43c0-97e3-3d4c5ca9de5b – 51:11
Comparing arc length values at the critical point and endpoint to ensure our result is indeed a maximum – 53:05
At α ≈ 0.9855 radians ≈ 56° the arc length is at a maximum and is approximately 1.20v²/g – 58:07
Calculation check – 1:00:12