A particle is projected at an angle of 60° with the horizontal from the ground with a velocity 10 √3 ms -1 . The angle between velocity vector after 2 s and initial velocity vector is (g=10 ms -2)

Question

A particle is projected at an angle of 60° with the horizontal from the ground with a velocity 10 √3 ms -1 . The angle between velocity vector after 2 s and initial velocity vector is (g=10 ms -2) (A) 0° (B) 30° (C) 60° (D) 90°

Tardigrade Admin · Accepted Answer

Initial velocity, v i =2 cos θ hat i +4 sin θ hat j =5 √3 hat i +15 hat j Final velocity vector (after 2 s ), v f =u cos θ hat i +(u sin θ-g t) hat j =5 √3 hat i -5 hat j Now, v i ⋅ v f =25 × 3-15 × 5=0 ∴ v i perp v f

Q. A particle is projected at an angle of $6 0^{\circ}$ with the horizontal from the ground with a velocity $103 m s^{- 1} .$ The angle between velocity vector after $2 s$ and initial velocity vector is $(g = 10 m s^{- 2})$

$0^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$

Initial velocity, $v_{i} = 2 cos θ \hat{i} + 4 sin θ \hat{j} = 53 \hat{i} + 15 \hat{j}$
Final velocity vector (after $2 s)$ ,
$v_{f} = u cos θ \hat{i} + (u sin θ - g t) \hat{j} = 53 \hat{i} - 5 \hat{j}$
Now, $v_{i} \cdot v_{f} = 25 \times 3 - 15 \times 5 = 0$
$∴ v_{i} ⊥ v_{f}$

Q. A particle is projected at an angle of 60∘ with the horizontal from the ground with a velocity 103​ms−1. The angle between velocity vector after 2s and initial velocity vector is (g=10ms−2)

0∘

30∘

60∘

90∘