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A particle is thrown horizontally with s...

A particle is thrown horizontally with speed 10m/s along the rim of a smooth fixed cylinder of height 20m. Taking g=`10m//s^(2)`, the time taken by the particle to reach the bottom assuming it to be always in contact with the cylinder is (in sec)

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`t=sqrt((2h)/(g))=sqrt((2xx20)/(10))=2s`
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