A particle is dropped from height 3R from surface of earth. Then, the velocity of particle at height R is (given R is radius of earth, g is acceleration due to gravity on the earth’s surface) :

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gR2

23gR

gR2

Path of particle is parabolic

answer is C.

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Detailed Solution

Principle of conservation of energy is to be applied. Since acceleration is variable −GmM4R=−GmM2R+12mv2v=GM2R=gR2

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