A particle of mass m is projected with 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.

Explanation

As shown in the figure, required torque `vectau=vecrxxvecF`. But as it is the product of force and the length of perpendicular `OC` drawn from the point of projection on the line of action of the force.
Here `OC=("Range")/(2)=(2u^(2)sin theta*cos theta)/(2g)`
so, torque `=("Force")xx(OC)`
`=mg((2u^(2)sin theta*cos theta)/(2g))=mu^(2)sin theta*cos theta`