JEE Main 2023 · 11 April · evening · Single correct · +4 / −1
A projectile is projected at 30° from horizontal with initial velocity 40 ms⁻¹. The velocity of the projectile at t=2 s from the start will be : (Given g=10 m / s² )
- (a) 20 √3 ms⁻¹
- (b) Zero
- (c) 20 ms⁻¹
- (d) 40 √3 ms⁻¹
Answer and solution
Answer (a) 20 √3 ms⁻¹
Solution
Physics sir · 3 steps
At t = 2 s the vertical velocity has just died out, so only the horizontal part is left.
- Split the velocity. Horizontal: uₓ = 40 cos 30° = 20√3 m s⁻¹. Vertical: u_y = 40 sin 30° = 20 m s⁻¹.
- Horizontal stays fixed. Gravity acts only downward, so nothing changes vₓ. It stays 20√3 m s⁻¹ for the whole flight.
v_y = u_y - gt = 20 - 10 × 2 = 0
- So. At t = 2 s the projectile is at the top of its path, and its velocity is just vₓ = 20√3 m s⁻¹, option (a).
The idea to keep. Treat the two directions separately: vₓ never changes, and only v_y feels gravity. At the highest point v_y = 0, but the projectile is still moving at u cos θ.
Answer from the NTA final answer key. Solution by PrepWiser.