5. Each leg of the tripod is 5 units long, and the angles between each pair of legs are equal and fixed. When it is placed on the ground, the height from the top to the ground is 4 units. Later, one leg is damaged, and 1 unit is cut off from the bottom. When it is placed on the ground again, the height from the top to the ground becomes
5. $\frac{144}{\sqrt{1585}}$.

As shown in Figure 5, the original tripod, when erected, forms a regular tetrahedron $P-ABC$, with side edges of length 5 and height $PO=4$. Therefore, the radius of the circumcircle of the base equilateral triangle is
$$
\begin{array}{l}
\frac{1}{\sqrt{3}} AB=\sqrt{5^{2}-4^{2}} \\
=3 .
\end{array}
$$

Thus, $AB=3 \sqrt{3}$.
Since the volume of the regular tetrahedron $V_{P-1}$ is
$$
V_{1}=\frac{1}{3} \times 4 \times \frac{\sqrt{3}}{4}(3 \sqrt{3})^{2}=9 \sqrt{3} .
$$

Let the top angle of the side face $\angle APB=\theta$.
By the cosine rule, we have
$$
\cos \theta=\frac{5^{2}+5^{2}-(3 \sqrt{3})^{2}}{2 \times 5 \times 5}=\frac{23}{50} .
$$

After cutting off $AD=1$, the volume of the resulting tetrahedron $P-DBC$ is $V_{2}=\frac{4}{5} V_{1}=\frac{36 \sqrt{3}}{5}$.
The length of the base sides is
$$
DB=DC=\sqrt{5^{2}+4^{2}-2 \times 5 \times 4 \times \frac{23}{50}}=\sqrt{\frac{113}{5}},
$$

The height from $D$ to the base side $BC$ is
$$
\sqrt{\frac{113}{5}-\left(\frac{3 \sqrt{3}}{2}\right)^{2}}=\frac{\sqrt{317}}{2 \sqrt{5}} \text {; }
$$

The area of the base $\triangle DBC$ is $S_{\triangle DBC}=\frac{1}{2} \times \frac{\sqrt{317}}{2 \sqrt{5}} \times 3 \sqrt{3}=\frac{3 \sqrt{951}}{4 \sqrt{5}}$.
Therefore, the distance from point $P$ to the base is $\frac{3 V_{P-1}}{S_{\triangle DBC}}=$
$$
\frac{144}{\sqrt{1585}}
$$