The correct option is A y=2x
We know that, the maximum height formula is,
Hmax=u2sin2θ2g ...(ii), and the range of the projectile is given by,
R=u2sin2θg ...(ii)
It is also given that the range of the projectile is double the maximum height reached
R=2H
u2sin2θg=2×u2sin2θ2g, by simplifying we get,
⇒tanθ=2
⇒vyvx=2=yx⇒y=2x
Hence option A is the correct answer