Five, the lateral faces of the pyramid $P-ABCD$ are all isosceles right triangles with legs of length 1, where $\angle APD=$ $\angle APB=\angle PBC=\angle PDC=90^{\circ}$. Find the height of the pyramid.
Quadrilateral $ABCD$ is a kite $(AB=AD, BC=CD)$ as shown in Figure 53(a). Therefore, $AC$ is its axis of symmetry.
Since $\angle PAB = \angle PAD = 45^{\circ}$, the plane $PAC$ is the symmetry plane of the quadrilateral pyramid $P-ABCD$. In the plane $PAC$, draw $PH \perp AC$, with $H$ as the foot of the perpendicular, then $PH$ is the height of the quadrilateral pyramid.

Let the intersection of $AC$ and $BD$ be $O$. Since $BD$ is perpendicular to the symmetry plane $PAC$, we have
$$
\angle BOC = \angle BOP = 90^{\circ}
$$

Given $PB = BC$, it follows that $CO = PO$.

Since $AP$ is perpendicular to $PB$ and $PD$, $AP \perp OP$.
In Figure 53(b), $PA = 1, PC = \sqrt{2}$. Let $\angle PCA = \alpha, PH = h$. Since $CO = PO$, $\angle CPO = \alpha$. Thus, $\angle POA = 2\alpha$,
$$
h = \sqrt{2} \sin \alpha = 1 \cdot \sin (90^{\circ} - 2\alpha) = \cos 2\alpha = 1 - 2 \sin^2 \alpha
$$

That is,
$$
h = 1 - h^2
$$

Therefore,
$$
h = \frac{-1 + \sqrt{5}}{2}.
$$