Prove that the maximum horizontal range is four times the maximum height attend by a projectile when fire at an inclination so as to have maximum horizontal range

Answer 1

Θ = launch angle (above horizontal)

v = launch velocity

flight time = 2 [v sin(Θ)] / g

horizontal velocity = v cos(Θ)

range = horizontal velocity * flight time
... = {[v cos(Θ)] * 2 [v sin(Θ)]} / g
... = v^2 sin(2Θ) / g

max sine occurs at π/2
... so max range occurs at π/4

max range = v^2 / g

max height = average velocity * time
... = [v sin(π/4)]^2 / (2 g)
... = v^2 / (4 g)