Angular momentum of projectile.

[Satvik Pandey]

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[/B]

 

[Doc Al]

Your answer looks correct. Perhaps a typo in the problem statement or answer sheet.

 

[Satvik Pandey]

Your answer looks correct. Perhaps a typo in the problem statement or answer sheet.

Thank you Doc Al.