1.

A particle is released freely from a height H. At a certain height, its kinetic energy is two times of its potential energy. Then, the height and the speed of the particle at that instant are respectively (g = acceleration due to gravity)

Answer»

`H/3,sqrt((2gH)/3)`
`H/3,2sqrt((GH)/3)`
`(2H)/3,sqrt((2gH)/3)`
`(H)/3,sqrt(2gH)`

Solution :
If particle FALLS by a distance X, then
`KE=1/2mv^(2)=1/2m(2GX)=mgx`
`PE=mg(H-x)`
As, `KE=2(PE)rArrmgx=2mg(H-x)rArrx=(2H)/3`
So, height of particle is `H-x=H/3`.
Speed of particle at `H/3` distance = `sqrt(2g((2H)/3))=2sqrt((gH)/3)`


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