Question

A particle of mass m is projected with a velocity v making an angle of 45$^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

• A. zero
• B. $mv^3 (4 \sqrt 2 g )$
• C. $ mv^3 (\sqrt 2 g ) $
• D. $ m \sqrt {2 g h^3} $

Answer 1

Answer: (B)

$ L = m \frac{v }{\sqrt 2 } r_1$
Here,$ r_1 = h = \frac{v^2 sin ^2 45 ^ \circ }{ 2g} = \frac{ v^2}{4g} $
$ \therefore \, \, \, \, \, \, \, \, L = m \bigg( \frac{v}{ \sqrt 2} \bigg) \bigg( \frac{v^2}{4g} \bigg) = \frac{ mv^3}{ 4 \sqrt {2g}}$