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: (a). vx = 40cos 30° = 20√(3) m/s (constant) vy = 40sin 30° - gt = 20 - 10 × 2 = 0 So v = 20√(3) m/s (the projectile is at the top), answer (a).

A projectile is projected at 30° from horizontal with initial… | JEE Main 2023 (11 Apr, Shift 2)

Q10JEE Main 2023 · 11 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight

A projectile is projected at $30^\circ$ from horizontal with initial velocity $40\text{ ms}^{-1}$ . The velocity of the projectile at $t = 2\text{ s}$ from the start will be: (Given $g = 10\text{ m/s}^{2}$ )

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Answer: (a)

$v_x = 40\cos 30^\circ = 20\sqrt{3}\text{ m/s}$ (constant)
$v_y = 40\sin 30^\circ - gt = 20 - 10 \times 2 = 0$
So $v = 20\sqrt{3}\text{ m/s}$ (the projectile is at the top), answer (a).

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