A particle of mass m is projected with a velocity v making an angle of `45^@` with the horizontal. The magnitude of the angular momentum of the projectile abut the point of projection when the particle is at its maximum height h is.

Options

A. zero
B. `(mvh^(2))/(sqrt2)`
C. `(mv^(2)h)/(2)`
D. `(mvh)/(sqrt2)`

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)/(sqrt2)`
Angular momentum of the projectile about the point of projection `=mv'h=m(v)/(sqrt2)j=(mvh)/(sqrt2)`