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 ux2+uy2=vx2+vy2+2gh,ux=vx uy2=vy2+2gh or   uy2=(2)2+2×10×0.4=12 uy=12=23 ms-1,ux=vx=6 ms-1 tanθ=uyux=236=13  ⇒  θ=30°