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Angular momentum of projectile.

  1. Sep 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Find out the angular momentum of particle as it reaches the highest point about the origin in projectile?????

    m=1 kg angle(i)=60 degree(with horizontal) u=10 m/s g=10


    2. Relevant equations



    3. The attempt at a solution
    As angular momentum=##mrvsinB ##

    Here velocity at top=##ucos\theta ##

    As sinB=h/r


    h=rsinB

    Now h=##\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } ##

    So angular momentum=##\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } ## ##mucos\theta ##

    On putting values I got
    ##\frac { 100{ sin }^{ 2 }(60) }{ 2\times 10 } 10cos(60)##

    ##5{ sin }^{ 2 }(60)10cos(60)##

    ##\frac { 15 }{ 4 } \times \frac { 10 }{ 2 } =18.75##
    But the answer is 187.5
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your answer looks correct. Perhaps a typo in the problem statement or answer sheet.
     
  4. Sep 24, 2014 #3
    Thank you Doc Al.
     
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