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A particle of mass m has been thrown up ...

A particle of mass m has been thrown up with intial speed u making as angle `theta` with the horizontal. Find the torque of its weight about the point of projection when it just reached the highest point.

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The correct Answer is:
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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`
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