Two men stand back-to-back and walk in opposite directions for $40$ yards each. Each of them then turns left and walks another $40$ yards each. In yards, how far are the two men from one another?
If the two men have turned left after walking forward $40$ yards, they are at opposite corners of a square of side length $40$ yards. [asy]
pair A,B,C,D,E;
A=(0,0);
B=(40,0);
C=(40,40);
D=(0,40);
E=(0,00);
label("Front",B,E);
draw(A--B--C--D--A);
label("40",(A--B).midpoint,W);
label("40",(D--A).midpoint,N);
label("40",(D--C).midpoint,E);
[/asy] The side length of the square formed is the hypotenuse of a $45^\circ\text{-}45^\circ\text{-}90^\circ$ triangle with legs of length $40$. Thus, each diagonal is $40\sqrt{2}$ yards. Therefore, as the two men are $40\sqrt{2}$ yards apart at this point, they remain $40\sqrt{2}$ apart after they walk to the corners of the square. The final answer is then $2(40\sqrt{2}) = \boxed{80\sqrt{2}}$ yards.
Since each man walks $40$ yards in one direction and then $40$ yards in another direction, the two men effectively form the sides of a right triangle. The distance between them is equal to the length of the hypotenuse of this triangle.
By the Pythagorean Theorem, the length of the hypotenuse is given by:
$\sqrt{(40)^2 + (40)^2} = \sqrt{1600 + 1600} = \sqrt{3200} = \boxed{40\sqrt{2}}$ yards.