A particle of mass m is projected with a velocity v at an angle of 45^ with horizontal. When the particle is at its maximum height, the magnitude of its angular momentum about the point of projection is :

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Correct answer: B. m v^3 4 2 g

Maximum height H= v^2 ^2 45^ 2 g = v^2 2 g 1 2 = v^2 4 g Momentum of particle at the highest point p=m v 45^ =m v / 2 Angular momentum = pH = m v 2 v^2 4 g = m v^3 4 2 g

A particle of mass m is projected with a velocity v at an angle of 45^ with horizontal. When the particle is at its maximum height, the magnitude of its angular momentum about the point of projection is :

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