Q 11-06-101JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
A body of mass $m$ is projected with a speed $u$ making an angle of $45^\circ$ with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as $\dfrac{\sqrt2\,mu^3}{Xg}$. The value of $X$ is ______.
Numerical value type. Enter your answer.
Answer: 8
At the top, $v = u\cos45^\circ = \dfrac{u}{\sqrt2}$ (horizontal) and $H = \dfrac{u^2\sin^245^\circ}{2g} = \dfrac{u^2}{4g}$.
$$L = mvH = m\frac{u}{\sqrt2}\cdot\frac{u^2}{4g} = \frac{mu^3}{4\sqrt2\,g} = \frac{\sqrt2\,mu^3}{8g}$$
$X = 8$.
Solution by Sreeraj P, M.Sc Physics