1.

A particle is dropped from height H. At a point its kinetic energy is x times of its potential energy. Find the speed of the particle at that point -

Answer»

`[2gxH]^(1//2)`
`[(2G(X+1)H)/(x)]^(1//2)`
`[(2gH)/((x+1))]^(1//2)`
`[(2gxH)/((x+1))]^(1//2)`

SOLUTION :`K.E.=x(P.E.)`
`Or K.E.=((x)/(x+1))T.E.=((x)/(X+1))MGH`
`v=sqrt(2gH((x)/(x+1)))`


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