Question

A particle of mass m is projected with a 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

At the highest point total force acting on the particle is its weight acting downward Range of the particle

= $\left(\frac{u^{2}sin2\theta }{g}\right)$

Therefore force is at a distance l

$\Rightarrow \frac{\left(totalrange\right)}{2}=\frac{v^{2}sin2\theta }{2g}$

(From the initial point)

Therefore $\tau =F\times r=mg\times v^2\frac{sin2\theta }{2g}$

$(v=\text{initial velocity})$

$=m{v}^2\frac{sin2\theta }{2}$

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