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
 

3is a crest4 is a trough 2 is the amplitude 1 is the wavelength

Less force and energy was applied