Projectile is fired in such away that its horizantal range is equa to three times its maximum height. what is the angle of projectile

Using the following formulas:

Range = [ Vo^2 * sin 2*theta ]/g

Height = [ Vo^2 * sin theta^2 ]/g

Since the relation between Range and height is given: (R = 3H), the 2 equations can be equated in terms of Vo^2 so that the only remaining unknown variable left will be the angle. This is done as shown:

(g*Range) / (sin 2*theta) = Vo^2

(g*Height) / (sin theta^2) = Vo^2

(g*Range) / (sin 2*theta) = (g*Height) / (sin theta^2)

Applying Range = 3*Height:

(g*3*Height) / (sin 2*theta) = (g*Height) / (sin theta^2)

substituting given values while Height cancels out:

(9.8*3) / (sin 2*theta) = (9.8) / (sin theta^2)

Angle theta = 33.67 degrees