A bridge is built in the shape of a semielliptical arch. The bridge has a span of 140 feet and a maximum height of 40 feet. Choose a suitable rectangular coordinate system and find the height of the arch at a distance of 50 feet from the center. The height is about $\square$ feet. (Round to two decimal places as needed.)

Step 1 ：The question is asking for the height of a semielliptical arch at a distance of 50 feet from the center. The semiellipse is defined by the equation \((x/h)^2 + (y/k)^2 = 1\), where h is the horizontal semi-axis (half the total width of the bridge), k is the vertical semi-axis (the maximum height of the bridge), x is the horizontal distance from the center of the ellipse, and y is the height at that distance.

Step 2 ：We know that h = 140/2 = 70 feet and k = 40 feet. We need to find y when x = 50 feet.

Step 3 ：We can rearrange the equation of the ellipse to solve for y: \(y = k\sqrt{1 - (x/h)^2}\).

Step 4 ：We can then substitute the known values into this equation to find the height of the arch at a distance of 50 feet from the center.

Step 5 ：Substituting the values, we get h = 70, k = 40, and x = 50.

Step 6 ：Calculating the value of y, we get y = 27.994168488950606.

Step 7 ：Rounding to two decimal places, we get the final answer.

Step 8 ：Final Answer: The height is about \(\boxed{27.99}\) feet.