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Question

A particle is projected at time t=0 from a point P on the ground with a speed V0 at an angle of 45o to the horizontal. What is the magnitude of the angular momentum of the particle about P at time t=v0/g

A
mv2022g
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B
mv302g
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C
mv202g
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D
mv3022g
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Solution

The correct option is D mv3022g

ux=v0cosθ=v02uy=v0sinθgt=v02v0gg=v0[122]¯¯¯¯P=mv02^i+mv0[122]^j¯¯¯r=x^i+y^jx=uxt=v02.v0g=v202v0g=v202g

y=v0sinθt12gt2=v02v0g12gv20g2=v20g2[1212]=v20g2[2222]L=¯¯¯rׯ¯¯¯P

=[v202g+v20g2[2222]^j]×[mv02^i+mv0[122]^j]=mv3022g

¯¯¯¯L=mv3022g


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