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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 angular momentum of the projectile about the point of projection when the particle at its maximum height h is

A

zero

B

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

C

`mv_(0)^(2) // 4 sqrt(2)g`

D

`m sqrt(2 gh^(3))`

Text Solution

Verified by Experts

The correct Answer is:
D
Speed of the particle at the top = horizontal component of the speed of projection
`rArr" "v = v_(0) cos theta_(0) cos 45^(@) = (v_(0))/(sqrt(2))`
The angular momentum of the particle about O
= L = mvr sin `theta` where `theta` = angle between `vec(v) & vec(r)`
`rArr" "L = mv h" "(because r sin theta = h)`
`h = (v_(0)^(2) sin^(2) theta)/(2g) = (v_(0)^(2)sin^(2)45^(@))/(2g) = (v_(0)^(2))/(4g)`, we obtain
`L = (mv_(0)^(3))/(4 sqrt(2)g)`, putting the value of v and h, `v_(0) =sqrt((4gh))`
We obtain `L = m sqrt(2gh^(3))`
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