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Question

A particle is released from a height H. At a certain height, its kinetic energy is two times its potential energy.The height and speed of the particle at that instant are

A
H3,2gH3
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B
H3,2gH3
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C
2H3,2gH3
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D
H3,2gH
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Solution

The correct option is B H3,2gH3
Using law of conservation of energy,
Ki+Ui=Kf+Uf
let initial position be when particle is dropped from height H
Ui=mgH
Ki=0
Given Kf=2Uf
Ki+Ui=Kf+Uf
mgH=3Uf

Uf =13mgH and Kf=23mgH
Now using Kf=mv22
and speed of particle will be
v=2g(2H3)=2gH3
Now using Uf=mgh
h=H3

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