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A particle is projected vertivally upwar...

A particle is projected vertivally upwards from the surface of earth `(radius R_e)` with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is ……

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The correct Answer is:
`(1)`
According to law of conservation of total energy `(PE + KE)` on the surface of earth `= PE` at height h
`-(GMm)/(R ) + (1)/(2) ((1)/(2) m upsilon_(e)^(2)) = - (GMm)/((R + h))`
or `-(GMm)/(R ) + (1)/(4)m ((2GM)/(R )) = - (GMm)/((R + h))`
or `- (1)/(2) (GMm)/(R ) = - (GMm)/((R + h))`
or `R + h = 2R` or `h = R = nR` (as per question)
`:. n = 1`
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