A particle is projected vertically upwards with a velocity `sqrt(gR)`, where `R` denotes the radius of the earth and `g` the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

Options

A. `R_(e )`
B. `2R_(e )`
C. `(R_(e ))/(2)`
D. None of these

Answer

A

Explanation

The projected speed is less than escape speed `V_(e)=sqrt(2gR_(e))`. The body will not reach infinity.
At maximum height the kinetic energy becomes zero.
Using conservation of energy,
`(mV^(2))/(2)-(GM_(e)m)/(R_(e))=-(GM_(e)m)/(r)`
`implies (mgR_(e))/(2)-(GM_(e)m)/(R_(e))=-(GM_(e)m)/(r)(GM_(e)=gR_(e)^(2))`
`implies(mgR_(e))/(2)-mgR_(e)=-(gR_(e)^(2)m)/(r)implies(mgR_(e))/(2)=(gR_(e)^(2)m)/(r)`
`implies r=2R_(e)(r=R_(e)+h) therefore` The height attained = `R_(e)`