Question

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.

Answer 1

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`