Question

A particle of mass $m$ is projected with a speed $u$ at an angle $\theta$ with the horizontal. What is the torque of its weight about the point of projection when it first reached the highest point.

Answer 1

At the highest point, the total force acting on the particle is its weight acting downward.
Range of the particle will be

$R=\frac{u^2\sin 2\theta }{g}$

When the particle reaches the maximum height, it has covered half of its range.
Therefore, force is at a $\perp$ distance,

$\Rightarrow \frac{\left(totalrange\right)}{2}=\frac{u^2\sin 2\theta }{2g}$ ( From the initial point )

Therefore $\tau =F\times r$ ($\theta =$ angle of projection )

$=mg\times \frac{v^2\sin 2\theta }{2g}$ ($v=$ initial velocity )

$=\frac{m{v}^2\sin 2\theta }{2}=m{v}^2\sin \theta \cos \theta$.