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 about the point of projection, when the particle is at its maximum height h, is

Options

A. zero
B. `(mv^(3))/(4sqrt2g)`
C. `(mv^(3))/(sqrt2g)`
D. `msqrt(2gh^(3))`

Answer

B, D

Explanation

About a given point, angular momentum = moment of linear momentum
Height of a projectile= h

`thereforeh=(v^(2)sin^(2)theta)/(2g)=(v^(2)xxsin^(2)45^(@))/(2g)=(v^(2))/(4g)` ….(i)
At maximum height, the velocity of projectile
`vcostheta=vcos45^(@)=v/(sqrt2)`
`therefore` Momentum at highest point `=(mv)/(sqrt2)`
`therefore` Moment of momentum about point O`=(mv)/(sqrt2)xxh`
`therefore` Angular momentum about O =`(mvh)/(sqrt2)`
`thereforeL=(mvh)/(sqrt2)` ...(ii)
Put h from (i) in (ii), we get,
`L=(mv)/(sqrt2)xx((v^(2))/(4g))=(mv^(3))/(4sqrt2g)` ...(iii)
From (i), eliminate v in (iii)
Angular momentum `=(mv^(3))/(4sqrt2g)=(mxx(4gh)^(3//2))/(4sqrt2g)`
or `L=msqrt(2gh^(3))` ...(iv)
Options (b, d) represent the answers.