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. `(mvh^(2))/(sqrt(2))`
C. `(mv^(2)h)/(2)`
D. `(mvh)/(sqrt(2))`

Answer

D

Explanation

When a particle is projected with a speed v at `45^(@)` with the horizontal then, velocity of the projectile at maximum height, `v'=v cos 45^(@)=(v)/(sqrt(2))` Angular momentum of the projectile about the point of projection
`=" mv' h"`
`=m(v)/(sqrt(2))h=("mvh")/(sqrt(2))`