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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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We know that the value of escape velocity of a particle from the surface of earth is given by
`v_(e)=sqrt(2GM//R)`
So, the minimum kinetic energ needed for the particle to escape is given by
`1/2mv_(e)^(2)=1/2mxx(2GM)/R=(GMm)/R`
As per questions, the kinetic energy given to the particle is `(GMm//2R)`.
If `h` is the height attained by the particle from the surface of the earth the its poetntial energy is `-GMm//(R+h)`.
According to the law of conservation of total energy.
`KE+PE` on the surface of earth `=PE` at height `h`
`1/2(GMm)/R+((-GMm)/R)=-(GMm)/(R+h)impliesh=R`
or `(GMm)/(2R)=-(Gmm)/(R+h)impliesh=R`
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