The correct option is A y=2x−5x2
Initial velocity of projectile =^i+2^j⇒ux=1,uy=2
We know that
Range of a projectile, R=u2sin2θg=2(ucosθ)(usinθ)g⇒R=2uxuyg=2×1×210=25tanθ=uyux=21=2
Using equation of trajectory of a projectile,
y=xtanθ(1−xR)y=2x(1−5x2)=2x−5x2
Alternate Solution:
ux=1,uy=2⇒tanθ=uyux=2u=√u2x+u2y==√5
Equation of trajectory of a projectile
y=xtanθ−gx22u2cos2θ
=2x−10x22×5×(1√5)2
i.e. y=2x−5x2