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

Answer

C

Explanation

The horizontal component of the velocity of projection remains unchanged throughout the time of flight
`v_(x)=vcos45^(@)=(v)/(sqrt(2))`
Magnitude of angular momentum of particle about its point of projection when particle is at highest point,
`L=mhv_(x)=(mhv)/(sqrt(2))(theta=45^(@))`,
The maximum height, `H=(v^(2)sin^(2)theta)/(2g)=(v^(2))/(4g)(theta=45^(@))`
Substituting, the value of height h, `L=(mv^(3))/(4sqrt(2g))`