At a height 0.4 m from the ground, the velocity of a projectile in vector form is v→=(6i^+2j^)ms−1. The angle of projection is
45∘
60∘
30∘
tan−1(3/4)
v2=u2−2gh or u2=v2+2gh
or ux2+uy2=vx2+vy2+2gh,ux=vx
So, uy2=vy2+2gh or uy2=(2)2+2×10×0.4=12uy=12=23ms−1,ux=vx=6ms−1tanθ=uyux=236=13⇒θ=30∘