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Question

# A stone of mass m is projected with a velocity u at an angle of 45o to the horizontal. Its angular momentum about the point of projection when it is at its highest point is :

A
mu342g
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B
mu24g
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C
mu2
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D
2mu3g
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Solution

## The correct option is A mu34√2gAt the highest point , velocity will only be in x directioni.e. →u=ucosθ^i+0^jwhile →r=rx^i+ry^jry=u2sin24502g,rx=Range2=u2sin9002gSo, Angular momentum about point of projection: →L=m(→r×→v)or, →L=m(rx^i+ry^j)×(ucosθ^i) or, →L=mryucosθ(−^k)=m(ucos450)(u24g) =mu34√2g.

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