jmc

If $x$ represents the length of the shorter leg, then the two legs are $x$ and $2x-1$. In a right triangle, the length of one leg is the base and the length of the other leg is the height, so the area of this triangle is $\frac{1}{2}bh=\frac{1}{2}x(2x-1)$. We set this equal to 60 and solve for $x$. $$\begin{array}{rl}\frac{1}{2}(2x^2-x)& =60\Rightarrow \\ 2x^2-x& =120\Rightarrow \\ 2x^2-x-120& =0\Rightarrow \\ (2x+15)(x-8)& =0\end{array}$$ Since $x$ must be positive, we get that $x=8$. The shorter leg is 8 feet long and the longer leg is $2(8)-1=15$ feet long. We can use the Pythagorean Theorem to find the hypotenuse or we recognize that 8 and 15 are part of the Pythagorean triple $8:15:17$. The hypotenuse of the right triangle is $\boxed{17}$ feet long.