At a height 0.4 m from the ground, the velocity of a projectile in vector form is: v→ = (6i^+2j^)m/s. The angle of projection is:

v2 = u2-2gh or u2 = v2+2ghor ux2+uy2 = vx2+v2y+2ghAs vx = ux;  uy2 = vy2+2ghor uy2 = (2)2+2 × 10 ×0.4 = 12uy = 12 = 23 m/sux = vx = 6 m/stan θ = uyux = 236 = 13θ = 300