The escape velocity of a body on the earth's surface is $v_e$. A body is thrown with a speed $3v_e$. Assuming that the sun and planets do not influence the motion of the body, its speed at infinity would be

If $v_i and v_f$are respectively the initial and final speeds of the body, we have, from the law of conservation of energy, 

$\frac{1}{2}m{v_i}^2-\frac{GmM}{R}=\frac{1}{2}m{v_f}^2 \left(i\right)$

where m is the mass of the body and M is the mass of the earth and R its radius. The escape velocity is given by 

$\frac{1}{2}m{v_e}^2=\frac{GmM}{R} \left(ii\right)$

Using (ii) in (i) gives 

$$\begin{gathered} \frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_e}^2=\frac{1}{2}m{v_f}^2 \\ or v_f={({v_i}^2-{v_e}^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \left(iii\right) \end{gathered}$$

Given $v_i=3v_e.$

Therefore, $v_f=2\sqrt{2}{v}_e.$