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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

`msqrt(2)gh^(3)`

Text Solution

Verified by Experts

b) The angular momentum is
`L = r xx p`
`L = rmvcosthetasintheta`
But, `H = rsintheta`
`L = mvHcostheta`
`=mv xx (v^(2)sin^(2)gtheta)/(2sqrt(2)g)`
But, `theta=45^(@)`
`therefore` `L = (mv^(3))/(4sqrt(2)g)`
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