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"^(-1)`
B. `sqrt(2) xx 11.2 "km s"^(-1)`
C. `sqrt(3) xx 11.2 "km s"^(-1)`
D. `sqrt(8) xx 11.2 "km s"^(-1)`

Answer

D

Explanation

Initial KE of the body `= (1)/(2)mv^(2)`
Here, `v=3xx11.2 "kms"^(-1)`
So, `KE=(1)/(2)xxm(3xx11.2)^(2)=9xx(1)/(2)mv_(e)^(2)`
As `(1)/(2)mv_(e)^(2)` energy is used up in coming out from the gravitational pull of the earth, so final KE should be `8 xx (1)/(2)mv_(e)^(2)`
Hence, `(1)/(2)mv'^(2)=8v_(e)^(2)`
`rArr v=sqrt(8)v_(e)=sqrt(8)xx11.2 "kms"^(-1)`.