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(D) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At the m

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(D) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At the maximum height, the velocity of the particle is purely horizontal, given by $v_x = u \cos 	heta$.The position vector at maximum height has a vertical component equal to the maximum height $H = \frac{u^2 \sin^2 	heta}{2g}$.The magnitude of angular momentum at maximum height is $L = m v_x H = m (u \cos 	heta) \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying this, we get $L = \frac{m u^3}{2g} \sin^2 	heta \cos 	heta$.For $	heta = 0^{\circ}$, $L = 0$. For $	heta = 90^{\circ}$ (or $\frac{\pi}{2}$), $L = 0$.Between $0$ and $\frac{\pi}{2}$, the function $f(	heta) = \sin^2 	heta \cos 	heta$ is positive and reaches a maximum value. This corresponds to the shape shown in graph $D$.

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