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(D) The angular momentum $L$ about the point of projection is given by $L = m(\vec{r} 	imes \vec{v})$.At the highest point, the velocity is $v = u \c

Vedclass · Accepted Answer

$\frac{mu^3 \sin 	heta \sin 2	heta}{4g}$

Vedclass · Answer

(D) The angular momentum $L$ about the point of projection is given by $L = m(\vec{r} 	imes \vec{v})$.At the highest point, the velocity is $v = u \cos 	heta$ (horizontal direction).The perpendicular distance from the point of projection to the velocity vector is the maximum height $H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$.Thus, $L = m(u \cos 	heta) 	imes H_{\max}$.Substituting $H_{\max}$, we get $L = m(u \cos 	heta) \left( \frac{u^2 \sin^2 	heta}{2g} ight) = \frac{m u^3 \cos 	heta \sin^2 	heta}{2g}$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$, we can write $\cos 	heta \sin 	heta = \frac{\sin 2	heta}{2}$.Substituting this into the expression: $L = \frac{m u^3 \sin 	heta}{2g} \left( \frac{\sin 2	heta}{2} ight) = \frac{m u^3 \sin 	heta \sin 2	heta}{4g}$.

The angular momentum of a projectile projected at an angle $\theta$ with the horizontal with speed $u$ about the point of projection when it is at the highest point of its trajectory is

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