Question

A particle of mass m is projected with a speed u at an angle θ 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 id its weight acting downward.
Range of the particle = u² sin 2π / g
Therefore force is at a ⊥ distance, = (total range)/2 = (v² sin 2θ)/2g
(From the initial point)
Therefore π = F × r ( θ = angle of projection)
= mg × v² sin 2θ/2g (v = initial velocity)
= mv² sin 2θ / 2 = mv² sinθ  cosθ