Question

A particle of mass \( m \) is projected with a velocity \( v \) making an angle of \( 30^\circ \) 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 –

• A. \( \sqrt{3} \frac{m v^2}{g} \)
• B. zero
• C. \( \frac{m v^3}{2g} \)
• D. \( \sqrt{3} \frac{m v^3}{16 g} \)

Hint:

At the maximum height in projectile motion, the vertical velocity becomes zero, making the angular momentum zero.

Answer 1

The Correct Option is B

Step 1: Understand angular momentum at maximum height.
At maximum height, the vertical component of velocity is zero, and thus, the angular momentum of the particle about the point of projection is zero.
Step 2: Conclusion.
Hence, the angular momentum at maximum height is zero. Final Answer: \[ \boxed{0} \]