Q: A particle of mass m is projected with velocity v at an angle θ with the horizontal. Its angular momentum about the point of projection when it is at the highest point of its trajectory.

(a) $ \displaystyle \frac{m v^3 sin^2\theta cos\theta }{2 g} $

(b) $ \displaystyle \frac{m v^3 sin^2\theta cos\theta }{ g} $

(c) $\displaystyle \frac{m v^2 sin^2\theta cos\theta }{2 g} $

(d) $ \displaystyle \frac{m v^2 sin^2\theta cos\theta }{4 g} $

Ans: (a)