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(B) The torque $\vec{	au}$ about the point of projection $O$ is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.At the highest point,the horizontal di

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$m u^2 \sin 	heta \cos 	heta$

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(B) The torque $\vec{	au}$ about the point of projection $O$ is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.At the highest point,the horizontal distance from the point of projection is $x = R/2$,where $R$ is the horizontal range.The horizontal range is $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.So,the horizontal distance at the highest point is $x = \frac{R}{2} = \frac{u^2 \sin 	heta \cos 	heta}{g}$.The force acting on the particle is its weight $\vec{F} = m\vec{g}$,which acts vertically downwards.The magnitude of the torque is $	au = F \cdot x = (mg) \cdot \left( \frac{u^2 \sin 	heta \cos 	heta}{g} ight)$.Simplifying this,we get $	au = m u^2 \sin 	heta \cos 	heta$.

$A$ particle of mass $m$ is thrown with an initial speed $u$ at an angle $\theta$ with the horizontal. Find the torque of its weight about the point of projection when it reaches the highest point.

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