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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**

**[/B]**

h=rsinB

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

So angular momentum=

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 187.5On 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