A particle of mass m is projected with a velocity 'u' making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :

Solution

Physics sir · 4 steps

At the top the velocity is purely horizontal, so L is just mass times horizontal speed times the height.

1. Velocity at the top. At maximum height the vertical velocity is zero. Only the horizontal part stays: vₓ = u cos 30° = √3/2u.
2. Find the height. The maximum height is h = (u² sin² 30°)/2g = (u² (1/4))/2g = u²/8g.
3. Lever arm. The momentum mvₓ points horizontally, and the particle is at height h above the launch point. So the perpendicular distance from the point of projection to the line of motion is h.

   L = m vₓ h = m × √3/2u × u²/8g = √3/16mu³/g

4. So. The angular momentum at the highest point is √3/16mu³/g, which is option (d).

The idea to keep. Angular momentum of a moving point is momentum times the perpendicular distance to its line of motion. For a horizontal velocity, that distance is just the height.

Answer from the NTA final answer key. Solution by PrepWiser.