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. `mv^(3)//4sqrt2g`
C. `mv^(2)//sqrt2g`
D. `m(2gh^(3))`

Answer

B

Explanation

Maximum height attained `(h)=(v^(2)sin^(2)theta)/(2g)`
`=(v^(2)sin^(2)45^(@))/(2g)=(v^(2))/(4g)" "(therefore sin 45^(@)=(1)/(sqrt2))`
At highest point, momentum = `mv co s45^(@)=(mv)/(sqrt2)`
`therefore` Angular momentum of the particle at the maximum height about on axis of projection
`=(mv)/(sqrt2)xx(v^(2))/(4g)=(mv^(3))/(4sqrt2g)`