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

v2=u2−2gh  or     u2=v2+2gh or  ux2+uy2=vx2+vy2+2gh,ux=vxSo, uy2=vy2+2gh  or  uy2=(2)2+2×10×0.4=12uy=12=23ms−1,ux=vx=6ms−1tan⁡θ=uyux=236=13⇒θ=30∘