A particle of mass ‘m’ is projected with a velocity ν making angl

1.	A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.
Answer» Answer is (b) mν³/4√2g Velocity at the highest point = ν cos 45° = ν/√2 Maximum height h = ν² sin² 45/2g = ν²/4g. L = angular momentum = [m(ν/√2) x (ν²/4g) = mν³/4√2g Since ν² = 4gh. Therefore L = m(2gh³)^(1/2). Which is not give.

A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.

Answer»

Answer is (b) mν³/4√2g

Velocity at the highest point = ν cos 45° = ν/√2

Maximum height h = ν² sin² 45/2g = ν²/4g.

L = angular momentum = [m(ν/√2) x (ν²/4g) = mν³/4√2g

Since ν² = 4gh.

Therefore L = m(2gh³)^(1/2). Which is not give.

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A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.

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