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A particle of mass m is projected with a...

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.

A

zero

B

`(mv^(3))/(4sqrt(2g))`

C

`(mV^(3))/sqrt(2g)`

D

`(mV)/(sqrt(2gh^(3)))`

Text Solution

Verified by Experts

The correct Answer is:
B, D
Angular momentum `=` (momentum)`xx`(perpendicular distance of the line of action of momentum from the axis of rotation)

Angular momentum about `O`, L`=mv/sqrt2xxh`…….i
`now h=(V^(2)sin^(2)theta)/(2g)=V(^(2))/(4g) [:'theta=45^@]`..........ii
From eqn i and ii we get
`L=(mv)/sqrt(2)(2sqrt(gh))h=msqrt(2gh^(3))`
also from eqn i and ii we get
`L=(mv)/sqrt(2)x(V^(2))/(4g)=(mV^(2))/(4sqrt2g)`
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