The escape velocity of a body from the earth is `11.2 km//s`. If a body is projected with a velocity twice its escape velocity, then the velocity of the body at infinity is (in `km//s`)

Explanation

Let u = velocity of projection of the body
v = velocity of the body at infinite distance
` therefore 1/2 mu^2 = 1/2 mv_e^2 + 1/2 mv^2`
when `v_e` = velocity of escape
or `v^2 = u^2 - v_e^2 " or " v = sqrt(u^2 - v_e^2) = sqrt((22.4)^2 - (11.2)^2)`
` = (11.2) sqrt(4 - 1) = 11.2 xx sqrt3`
` therefore ` velocity of body at infinity ` = 11.2 sqrt3 km s^(-1)`