Sayan Asked a Question
June 6, 2020 9:47 pmpts 30 pts
A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is (b) mv'/(42 g) (d) my2gh (a) Zero (c) mv'/(V2 g) n into two narticles
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  • Ruby negi thankyou
    note that the direction of angular momentum is along negative z-direction...
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    Ruby negi
    use right hand rule...
  • Ruby negi Best Answer
    see the solution, if u have any doubt, let me know...
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    Ruby negi
    for any doubt