A particle of mass m is projected with velocity v making an angle of 45 with horizontal. Magnitude of angular momentum of particle about the point of projection when the particle is at its maximum height is

Options

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

Answer

D

Explanation

At highest point A, `v_(x)=vcostheta=vcos45^@=v/sqrt(2)`
And maximum height, `H=(v^(2)in^(2)45^@)/(2g)=v^(2)/(4g)`
`:.` Angular momentum= Linear momentum `xxbot` distance
`L=(mv)/sqrt(2)xxv^(2)/(4g)=(mv^(3))/(4sqrt(2))g`