Answer
$-\frac{5}{6}$
Work Step by Step
By definition, the distance between any point on the unit circle and the origin is 1. In this case, the distance between $(\frac{\sqrt{11}}{6}, y)$ and $(0, 0)$ is 1. Using the distance formula:
$\sqrt{(\frac{\sqrt{11}}{6}-0)^2+(y-0)^2}=1$
$\sqrt{(\frac{\sqrt{11}}{6})^2+y^2}=1$
$\sqrt{\frac{11}{36}+y^2}=1$
$\frac{11}{36}+y^2=1$
$y^2=\frac{25}{36}$
$y=\pm\frac{5}{6}$
Since it's given in the problem that the point is in Quadrant IV, where y is negative, the answer is $-\frac{5}{6}$.
