A particle of mass 'm' is projected with a velocity `upsilon` making an angle of `30^(@)` with the horizontal. The magnitude of angular moment of the projectile about the point of projection when the particle is at its maximum height 'h' is

Options

A. `(sqrt(3))/(2)(m upsilon^(2))/(g)`
B. zero
C. `(m upsilon^(3))/(sqrt(2)g)`
D. `(sqrt(3))/(2)(m upsilon^(2))/(g)`

Answer

D

Explanation

Max. height, `H = (upsilon^(2) sin 30^(@))/(2g)`
Momentum at the highest position `= mv cos 30^(@)`
`:.` Angular momentum of particle
`L = mv cos 30^(@) xx (upsilon^(2) sin^(2) 30^(@))/(2g)`
`= m upsilon xx (sqrt(3))/(2) xx (upsilon^(2) (1//2)^(2))/(2g)`
`= (sqrt(3))/(16 g) m upsilon^(3)`