We are given that the bridge is shaped like a parabolic arch, with a horizontal distance of 26 feet from the center to the edge (making the total span 52 feet), and the highest point (the vertex of the parabola) is 5 feet above the ground.

We can model this problem using the equation of a parabola in the form:

$y = ax^2 + bx + c$

Since the parabola is symmetric and its highest point is at the center (vertex), we can simplify it to:

$y = ax^2 + c$

At the highest point, the vertex, which is at $x = 0$, the height is 5 feet. Therefore, we have:

$y(0) = 5 \quad \Rightarrow \quad c = 5$

Next, we know that at a distance of 26 feet from the center (at $x = 26$), the height of the arch is 0 feet (since it's touching the ground). So we use this information to find $a$:

$y(26) = 0 \quad \Rightarrow \quad 0 = a(26)^2 + 5$

Solving for $a$:

$0 = 676a + 5$ $676a = -5$ $a = -\frac{5}{676}$

Now, we have the equation of the parabola:

$y = -\frac{5}{676}x^2 + 5$

To find the height of the bridge at a point 5 feet from the center ($x = 5$):

$y(5) = -\frac{5}{676}(5)^2 + 5$ $y(5) = -\frac{5}{676}(25) + 5$ $y(5) = -\frac{125}{676} + 5$ $y(5) = 5 - 0.1849$ $y(5) \approx 4.8151 \, \text{feet}$

So, the height of the bridge at a point 5 feet from the center is approximately 4.82 feet.

Would you like more details on any step or have any further questions?

Here are 5 questions for further exploration:

1. What is the total width of the bridge?
2. How would the equation change if the height of the arch increased?
3. What would be the height 10 feet from the center?
4. How does the shape of the parabola affect the design of the bridge?
5. What are other real-world examples of parabolic arches?

Tip: When working with parabolas, remember that the symmetry about the vertex can often simplify calculations!