Question

A particle of mass m is projected with a velocity v at an angle of $60^{\circ}$ with horizontal. When the particle is at its maximum height. The magnitude of its angular momentum about the point of projection is

• A. zero
• B. $ \frac{ 3 mv^2 }{ 16 g } $
• C. $ \frac{ \sqrt 3 mv^3 }{ 16 g } $
• D. $ \frac{ 3 mv^2 }{ 8 g } $

Answer 1

Answer: (B)

Maximum height, $H = \frac{ v^2 \, \sin^2 \, 60^\circ }{ 2 g } $
$= \frac{ v^2 }{ 2 g } \times \frac{ 3}{ 4} = \frac{ 3 v^2 }{ 8 g } $
Momentum of particle at highest point $p = mv \cos 60^\circ = \frac{ mv }{ 2} $
Angular momentum = pH
$= \frac{ mv }{ 2} \times \frac{ 3 v^2 }{ 8 g } $
$= \frac{ 3 mv^3 }{ 16 g } $