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(C) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p}$.At the maximum height, the velocity o

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$u^3$

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(C) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p}$.At the maximum height, the velocity of the particle is purely horizontal, given by $v_x = u \cos 	heta$.The position vector of the particle at maximum height is $\vec{r} = x \hat{i} + h_{\max} \hat{j}$, where $h_{\max} = \frac{u^2 \sin^2 	heta}{2g}$.The momentum is $\vec{p} = m v_x \hat{i} = m u \cos 	heta \hat{i}$.The angular momentum is $\vec{L} = (x \hat{i} + h_{\max} \hat{j}) 	imes (m u \cos 	heta \hat{i}) = -m u \cos 	heta h_{\max} \hat{k}$.The magnitude is $L = m u \cos 	heta \left( \frac{u^2 \sin^2 	heta}{2g} ight) = \frac{m u^3 \sin^2 	heta \cos 	heta}{2g}$.Thus, $L \propto u^3$.

$A$ body of mass $m$ is thrown with velocity $u$ from the origin of a coordinate system at an angle $\theta$ with the horizontal. The magnitude of the angular momentum of the particle about the origin at the time $t$ when it is at the maximum height of the trajectory is proportional to

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