A particle of mass $$m$$ 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 :

A particle of mass $$m$$ is projected with velocity $$u$$ at angle $$30°$$ with the horizontal. We need the angular momentum about the point of projection at maximum height.

Find velocity and position at maximum height: at maximum height, only the horizontal component of velocity remains:

$$v_x = u\cos 30° = \frac{u\sqrt{3}}{2}$$

The maximum height is:

$$h = \frac{u^2 \sin^2 30°}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g}$$

Compute angular momentum: the angular momentum formula about the point of projection is $$L = m \cdot v_x \cdot h$$ (since the velocity is purely horizontal and the perpendicular distance from the projection point to the velocity line is $$h$$).

$$ L = m \cdot \frac{u\sqrt{3}}{2} \cdot \frac{u^2}{8g} = \frac{\sqrt{3}}{16} \cdot \frac{mu^3}{g} $$

The correct answer is Option (1): $$\frac{\sqrt{3}}{16} \frac{mu^3}{g}$$.