Q 11-06-096JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
A particle of mass $m$ is projected with a velocity $u$ 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:
Answer: (A) $\dfrac{\sqrt3}{16}\dfrac{mu^3}{g}$
At the highest point the velocity is horizontal, $u\cos30^\circ$, and the particle is at height
$$h = \frac{u^2\sin^2 30^\circ}{2g} = \frac{u^2}{8g}$$
The lever arm of the horizontal momentum about the launch point is $h$:
$$L = m\,u\cos30^\circ \cdot h = m\cdot\frac{\sqrt3 u}{2}\cdot\frac{u^2}{8g} = \frac{\sqrt3}{16}\frac{mu^3}{g}$$
Solution by Sreeraj P, M.Sc Physics