A particle is projected vertically upwards the surface of the earth (radius `R_(e))` with a speed equal to one fourth of escape velocity what is the maximum height attained by it from the surface of the earth?

Options

A. `16/15 R_(e)`
B. `R_(e)/15`
C. `4/15 R_(e)`
D. None of these

Answer

B

Explanation

From conservation of mechanical energy, `1/2 mv^(2)=(GMm)/R_(e)- (GMm)/R`
Where R = maximum distance from centre of the earth Also `v=1/4 v_(e)=1/4 sqrt((2GM)/R_(e))`
`implies 1/2 mxx1/16xx(2GM)/R_(e)=(GMm)/R_(e)-(GMm)/R implies R=16/15 R_(e) implies h=R-R_(e)=R_(e)/15`.