The escape speed of a body on the earth's surface is `11.2kms^(-1)`. A body is projected with thrice of this speed. The speed of the body when it escape the gravitational pull of earth is

Explanation

Here, `upsilon_(e) = 11.2 km s^(-1)`. The velocity with which a projectile is projected is `upsilon = 3 upsilon_(e)`. Let `m` be the mass of the projectile and `upsilon_(0)` be the velocity of the projectile after escaping the gravitational field of earth. Using law of conservation of energy, we have
`(1)/(2) m upsilon_(0)^(2) = (1)/(2) m upsilon^(2) - (1)/(2) m upsilon_(e)^(2)`
or `upsilon_(0)^(2) = upsilon^(2) - upsilon_(e)^(2)`
or `upsilon_(0) = sqrt(upsilon^(2) - upsilon_(e)^(2)) = sqrt((3 upsilon_(e))^(2) - upsilon_(e)^(2)) = sqrt(8 upsilon_(e)^(2))`
`= 2 sqrt(2) upsilon_(e) = 2 xx (1.414) xx 11.2 km s^(-1)`.
`= 31.68 km s^(-1)`