The correct option is A 30∘
v=vx^i+vy^j=6^i+2^j
We know, in case of projectile motion, horizontal component of the velocity remains constant.
∴vx=ux=6
For y- component, using third equation of motion
v2y=u2y−2(10)(0.4)⇒u2y=12uy=√12
=2√3
The angle of projection θ is,
tanθ=uyuxtanθ=2√36θ=tan−11√3⇒θ=30∘