Engineering Mechanics: Statics & Dynamics (14th Edition)

$y=3.81ft$
The required maximum distance can be determined as follows: We know that $\Sigma U_{1-2}=12y-(2\times 15\times (t-4))=0$ As $t=\sqrt{y^2+(4)^2}$ $\implies 12y-30(\sqrt{y^2+(4)^2})=0$ $\implies 12y=30\sqrt{y^2+(4)^2}=0$ This simplifies to: $y=3.81ft$