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

A particle of mass W is projected with a velocity V making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height `H` is:

zero

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

`(sqrt(3)mv^(3))/(16g)`

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

Text Solution

Verified by Experts

The correct Answer is:

Here the angular momentain ‘L’ is given by
`L=pr=mvcostheta.H`
`=mvcos30^@xx(v^(2)sin^(2)30^@)/(2g)`
`mv^(3)xxsqrt(3)/2xx1/(4xx2g)=(sqrt(3)mv^(3))/(16g)`

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A projectile of mass 100 g is fired with a velocity of 20 m s^(-1) making an angle of 30^(@) with the horizontal. As it rises to the highest point of its path, its momentum changes by:

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1kg m`s^(-1)`

2kg m`s^(-1)`

None of these