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JEE Main 2026 (4 Apr, Shift 2) · Velocity, Momentum & Energy in Flight

Velocity, Momentum & Energy in Flight question from JEE Main 2026 (4 Apr, Shift 2)

Q2JEE Main 2026 · 4 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight

If xx and yy coordinates of a projectile as a function of time (t)(t) are given as 24t24t and 43.6t−4.9t243.6t - 4.9t^{2}, respectively, then the angle (in degrees) made by the projectile with horizontal when t=2 st = 2\text{ s} is ______.

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Answer: (b)

  1. vx=dxdt=24 m/sv_x = \frac{dx}{dt} = 24\text{ m/s}, vy=dydt=43.6−9.8tv_y = \frac{dy}{dt} = 43.6 - 9.8t
  2. At t=2 st = 2\text{ s}: vy=43.6−19.6=24 m/sv_y = 43.6 - 19.6 = 24\text{ m/s}
  3. tan⁡α=vyvx=2424=1⇒α=45∘\tan\alpha = \frac{v_y}{v_x} = \frac{24}{24} = 1 \Rightarrow \alpha = 45^\circ
  4. So the answer is (b) 45.
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