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A particle of mass m is projected with v...

A particle of mass m is projected with velocity v at an angle `theta` with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

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As shown in the figure, required torque `vec(tau)=vec(r )xx vec(F)`.
But as it is the product of force and the length of perpendicular OC drawn from the point of projection on the line of action of the force.
Here, `OC = ("Range")/(2)`
`=(2u^(2)sin theta.cos theta)/(2g)`
So, torque = (Force) `xx` (OC)
`= mg((2u^(2)sin theta.cos theta)/(2g))`
`= mu^(2)sin theta.cos theta`
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