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?

Explanation

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