jmc

To begin this problem, we first notice that the side length of this equilateral triangle is $8$ (the distance between the two points given). We then consider the altitude of an equilateral triangle with side length $8$. If we draw an equilateral triangle and its altitude, we notice that the altitude splits the equilateral triangle into two $30-60-90$ triangles with the side length being the hypotenuse of these triangles. In our case, the hypotenuse of the $30-60-90$ triangle has length $8$, so the altitude (the side opposite the $60^{\circ }$ angle) has length $4\sqrt{3}$ because the side lengths of a $30-60-90$ triangle are related in the ratio $1:\sqrt{3}:2$.

Since the base of the equilateral triangle is at a $y$-coordinate of $5$ and $4\sqrt{3}>5$, for the third vertex to be in the first quadrant, its $y$ coordinate must be $5$ $+$ the altitude of the equilateral triangle. The third vertex has a $y$-coordinate of $\boxed{5+4\sqrt{3}}$.