Question

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 1

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.