This problem is a Pythagorean Theorem problem, I assume?

- The rope is the hypotenuse (h = 26 yards)
- Length of the ground from the pole is side 'a' (a = 22 yards)

Thus, the set-up for this problem is:

$${{\mathtt{22}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{26}}}^{{\mathtt{2}}}$$

Taking these values, the next step is to square 22 and 26.

22^2 = 484

26^2 = 676

The equation after simplifying the numbers is:

$${\mathtt{484}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{676}}$$

Now, it is necessary to isolate the 'b^2'.

$$\left({\mathtt{484}}{\mathtt{\,-\,}}{\mathtt{484}}\right){\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = \left({\mathtt{676}}{\mathtt{\,-\,}}{\mathtt{484}}\right)$$

$${{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{192}}$$

Now, in order to completely isolate the 'b^2', we must take the square root of it, and in turn, 192.

$${\sqrt{{{\mathtt{b}}}^{{\mathtt{2}}}}} = {\sqrt{{\mathtt{192}}}}$$

$${\mathtt{b}} = {\sqrt{{\mathtt{192}}}}$$

The square root of 192 is 13.8564064605510183, which, when rounded to the nearest tenth, is about 13.9

@Anon: You rounded to the nearest hundredth, but you summed it up pretty well. I do have a habit of extracting it to the barebones