find the angle of projection at the horizontal range is twice the maximum height of a projecyile

Answer 1

Range is x = vo^2 sin2θ/g
Max Height is y = (vo Sinθ)^2/2g
Set range = twice max height
The vo^2 will cancel, solve for θ

Answer 2

find the angle of projection at which the horizontal range is twice the maximum teight of a projective.