Physics Circle home Student Login

Home›Chapters›Projectile Motion PYQs›Velocity, Momentum & Energy in Flight›Question 6

JEE Main 2025 (22 Jan, Shift 2) · Velocity, Momentum & Energy in Flight

Velocity, Momentum & Energy in Flight question from JEE Main 2025 (22 Jan, Shift 2)

Q6JEE Main 2025 · 22 JAN · SHIFT 2Easy

Velocity, Momentum & Energy in Flight

A ball of mass 100 g is projected with velocity 20 m/s20\text{ m/s} at 60∘60^\circ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

Show answer and solution

Answer: (c)

  1. Loss of KE up to the top equals the KE of the vertical component: ΔK=12m(usin⁡θ)2\Delta K = \frac{1}{2}m(u\sin\theta)^2.
  2. usin⁡60∘=20×32=103 m/su\sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ m/s}
  3. ΔK=12×0.1×300=15 J\Delta K = \frac{1}{2} \times 0.1 \times 300 = 15\text{ J}
  4. So the answer is (c) 15 J.
Report a mistake

More questions on Velocity, Momentum & Energy in Flight