Question

# At a height of 0.4 m from the ground, the velocity of a projectile is →v=(6^i+2^j) m/s . The angle of projection is (g=10 m/s2 ).45∘60∘30∘tan−1(34)

Solution

## The correct option is C 30∘Given v=vx^i+vy^j=6^i+2^j As the horizontal component of velocity always remains constant ∴vx=ux=6 By using third equation of motion; v2y=u2y−2gh 22=u2y−2(10)(0.4) ⇒u2y=12 or uy=√12=2√3 tanθ=uyux=1√3⇒θ=30° Hence, the correct answer is option (c).

Suggest corrections