A body is projected upwards with a velocity of `4 xx 11.2 "km s"^(-1)` from the surface of earth.What will be the velocity of the body when it escapes from the gravitational pull of earth ?

Options

A. 11.2 km/s
B. `sqrt(15)xx11.2` km/s
C. `2xx11.2` m/s
D. `10xx11.2` km/s

Answer

B

Explanation

Initial K.E. `=(1)/(2)mv^(2)=(1)/(2)m(4xx11.2)^(2)`
`=16xx(1)/(2)mv_(e )^(2)`
where `v_(e )=11.2 km//s`
When it escapes the gravitational field,
`K.E.=[16xx(1)/(2)mv_(e )^(2)]-(1)/(2)mv_(e )^(2)`
`(1)/(2)mv'^(2)=15[(1)/(2)mv_(e )^(2)]`
`therefore v'=sqrt(15)v_(e )=sqrt(15)xx11.2` km/s