JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Rotational Mechanics

At the highest point of projectile motion, the vertical velocity component is zero and the horizontal component is

$v_x=u\cos 45^{\circ }=\frac{u}{\sqrt{2}}$

The maximum height reached is

$h=\frac{u^2{\sin }^{2}45^{\circ }}{2g}=\frac{u^2}{4g}$

The angular momentum about the point of projection at the highest point is the product of linear momentum and the perpendicular distance (the height):

$L=m{v}_{x}h=m\times \frac{u}{\sqrt{2}}\times \frac{u^2}{4g}=\frac{m{u}^3}{4\sqrt{2}g}=\frac{\sqrt{2}m{u}^3}{8g}$

Comparing with $\frac{\sqrt{2}m{u}^3}{Xg}$, X = 8