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

A, B, C, D

Explanation

At the highest point total force acting on the particle is its weight acting downward Range of the particle
`=((u^2sin 2theta)/g)`
therefore forceis at a distance l
`rarr (total range)/2=((v^2sinw2theta))/(2g)`
`(from the initial point)
Therefore `tau=Fxxr=mgxv^2(sin2theta)/(2g)`
`(v=` initial velocity)
`mv^2(sin2theta)/2`
`=mv^2sintheta.costheta`