A cannonball is fired in such a way that its horizontal range is equal to 3
times its maximum height. Find the
launch angle of this cannonball.

θ = 53.13° above horizontal

Explanation:

Ignore air resistance.

The time the ball takes to reach its max height will equal the time to travel half the max range R.

√(2h/g) = (R/2)/vcosθ

vsinθ/g = R/2vcosθ

2v²sinθcosθ = Rg

2v²sinθcosθ = 3hg

v² = u² + 2gh

0² = v²sin²θ + 2gh

v²sin²θ = 2gh

v² = 2gh/sin²θ

               2v²sinθcosθ = 3hg

(2gh/sin²θ)2sinθcosθ = 3hg

            (2/sinθ)2cosθ = 3

                            cotθ = 3/4

                                 θ = 53.130

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grav potential energy