Let $\mathcal S$ be the sphere with center $(0,0,1)$ and radius $1$ in $\mathbb R^3$.  A plane $\mathcal P$ is tangent to $\mathcal S$ at the point $(x_0,y_0,z_0)$, where $x_0$, $y_0$, and $z_0$ are all positive.  Suppose the intersection of plane $\mathcal P$ with the $xy$-plane is the line with equation $2x+y=10$ in $xy$-space.  What is $z_0$?
1. **Determine the equation of the tangent plane:**
   The plane $\mathcal{P}$ is tangent to the sphere $\mathcal{S}$ at the point $(x_0, y_0, z_0)$. The normal vector to the plane at this point is $\langle x_0, y_0, z_0 - 1 \rangle$ because it is the gradient of the sphere's equation $(x - 0)^2 + (y - 0)^2 + (z - 1)^2 = 1$ at $(x_0, y_0, z_0)$.

   The equation of the plane $\mathcal{P}$ can be written as:
   \[
   x_0(x - x_0) + y_0(y - y_0) + (z_0 - 1)(z - z_0) = 0
   \]

2. **Simplify the plane equation:**
   Expanding and simplifying the equation:
   \[
   x_0 x - x_0^2 + y_0 y - y_0^2 + (z_0 - 1)z - (z_0 - 1)z_0 = 0
   \]
   \[
   x_0 x + y_0 y + (z_0 - 1)z = x_0^2 + y_0^2 + (z_0 - 1)z_0
   \]

3. **Intersection with the $xy$-plane:**
   When $z = 0$, the equation of the plane becomes:
   \[
   x_0 x + y_0 y = x_0^2 + y_0^2 + (z_0 - 1)z_0
   \]

4. **Use the given line equation:**
   The intersection of the plane $\mathcal{P}$ with the $xy$-plane is given by the line $2x + y = 10$. This implies:
   \[
   x_0 x + y_0 y = z_0
   \]
   Comparing with $2x + y = 10$, we get:
   \[
   x_0 = 2y_0 \quad \text{and} \quad z_0 = 10y_0
   \]

5. **Use the sphere equation:**
   The point $(x_0, y_0, z_0)$ lies on the sphere $\mathcal{S}$, so:
   \[
   x_0^2 + y_0^2 + (z_0 - 1)^2 = 1
   \]
   Substituting $x_0 = 2y_0$ and $z_0 = 10y_0$:
   \[
   (2y_0)^2 + y_0^2 + (10y_0 - 1)^2 = 1
   \]
   \[
   4y_0^2 + y_0^2 + (100y_0^2 - 20y_0 + 1) = 1
   \]
   \[
   105y_0^2 - 20y_0 + 1 = 1
   \]
   \[
   105y_0^2 - 20y_0 = 0
   \]
   \[
   y_0(105y_0 - 20) = 0
   \]
   Since $y_0 \neq 0$ (as $y_0$ is positive), we have:
   \[
   105y_0 = 20
   \]
   \[
   y_0 = \frac{20}{105} = \frac{4}{21}
   \]

6. **Find $z_0$:**
   \[
   z_0 = 10y_0 = 10 \times \frac{4}{21} = \frac{40}{21}
   \]

The final answer is $\boxed{\frac{40}{21}}$