Angular momentum of projectile.

  • #1
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Homework Statement


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


Homework Equations





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
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Answers and Replies

  • #2
Doc Al
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Your answer looks correct. Perhaps a typo in the problem statement or answer sheet.
 
  • Like
Likes Satvik Pandey
  • #3
591
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Your answer looks correct. Perhaps a typo in the problem statement or answer sheet.
Thank you Doc Al.
 

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