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. (`i`),(`ii`)
B. (`ii`),(`iii`)
C. (`ii`),(`iv`)
D. (`iii`),(`iv`)

Explanation

`l_(0)=mv cos45^(@)xxh`
` =mv cos45^(@)xx(V^(2) sin^(2) 45^(@))/(2g)`
`=(mv^(3))/(2g)*(1)/(2sqrt2)=(mv^(3))/(4sqrt2g)`
`h=(v^(2) sin^(2)45^(@))/(2g)=(v^(2))/(4g)implies v=sqrt(4gh)`
`l_(0)=(m)/(4sqrt2)(4gh)^(3//2)=msqrt(2gh^(3))`