A particle of mass m is projected with velocity v moving at an angle of `45^(@)` with horizontal. The magnitude of angular momentum of projectile about point of projection when particle is at maximum height, is

Options

A. zero
B. `(mv^(3))/(4sqrt(2g))`
C. `(mv^(3))/sqrt(2g)`
D. `m^(2)sqrt(2gh^(3))`

Answer

B

Explanation

`L=mu_(x)xxh`
`=(mv cos theta)xx((v sin theta)^(2))/(2g)=(mv^(3))/(2g) (cos theta sin^(2) theta)`
`=mv^(3)(cos 45^(@)sin^(2)45^(@))/(2g)=(mv^(3))/(2g)xx(1)/(2sqrt(2))=(mv^(3))/(4sqrt(2g))`