1.

A projectile of mass m is projected with a velocity v making an angle 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the harticle is at its maximum height h is :

Answer»

zero
`(mv^(3))/(SQRT(2)G)`
`m^(2)sqrt(2gh^(3))`
`(mv^(3))/(4sqrt(2)g)`

Solution :At highest point A, `v_(X)=vcostheta=vcos45^@=v/sqrt(2)`
And maximum height, `H=(v^(2)in^(2)45^@)/(2g)=v^(2)/(4g)`
`:.` Angular momentum= Linear momentum `xxbot`distance
`L=(mv)/sqrt(2)xxv^(2)/(4g)=(mv^(3))/(4sqrt(2))g`


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