In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, 
and side $BC=s$. The length of $BN$ is

$\textbf{(A)}\ s\sqrt 2 \qquad  \textbf{(B)}\ \frac 32s\sqrt2 \qquad  \textbf{(C)}\ 2s\sqrt2 \qquad  \textbf{(D)}\ \frac{s\sqrt5}{2}\qquad \textbf{(E)}\ \frac{s\sqrt6}{2}$
Suppose that $P$ is the intersection of $\overline{CM}$ and $\overline{BN}.$ Let $BN=x.$ By the properties of centroids, we have $BP=\frac23 x.$
Note that $\triangle BPC\sim\triangle BCN$ by AA. From the ratio of similitude $\frac{BP}{BC}=\frac{BC}{BN},$ we get
\begin{align*} BP\cdot BN &= BC^2 \\ \frac23 x\cdot x &= s^2 \\ x^2 &= \frac32 s^2 \\ x &= \boxed{\textbf{(E)}\ \frac{s\sqrt6}{2}}. \end{align*}
~MRENTHUSIASM