Question:
A particle of mass $m$ is projected with the speed $u$ at an angle $\theta$ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.
Solution:
$\begin{aligned} \tau &=F \times d_{\Perp} \\ &=m g\left(\frac{\operatorname{Range}}{2}\right) \\ &=m g\left(\frac{u^{2} \sin 2 \theta}{2 g}\right) \\ \tau &=m u^{2} \sin \theta \cos \theta \end{aligned}$
Torque acts perpendicular to plane of the motion.