#### Answer

$y=3.81ft$

#### Work Step by Step

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:
$y=3.81ft$