Aprojectile is fired in such a way that its horizontal range is equal to 14.5 times its maximum height. what is the angle of projection?

The maximum height is: hmax  = v0² sin² α / 2 g
The horizontal range is: x = v0² sin 2 α / g

So we are given that:

x = 14.5 hmax

v0² sin 2 α / g = 14.5 v0² sin² α / 2 g
7.25 sin² α = 2 sin α cos α    
7.25 sin α = 2 cos α  
7.25 tan α = 2
tan α = 0.27586
α = tan^(-1) 0.27586
α = 15.42°
The angle of projection is 15.42°

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