Will a truck that is 8 feet wide carrying a load that reaches 7 feet above the ground clear the semielliptical arch on the one-way road that passes under the bridge shown in the figure on the right? It (Round to two decimal places as needed.)

We begin by finding the height of the arch at the point where the truck first “meets” it. Since the truck is 8 feet wide its edge is \(4\) feet from the center. Assume that the semielliptical arch (the part above the roadway) has a horizontal semiaxis of \(a=9\) feet and a vertical semiaxis of \(b=10\) feet. (Thus the full width is \(18\) feet and the maximum height is \(10\) feet.) Then the equation of the ellipse is \[ \frac{x^2}{9^2}+\frac{y^2}{10^2}=1. \] Since we are interested only in the top half, \(y\ge0\). Step 1. Plug in \(x=4\) feet (half the truck’s width) into the equation: \[ \frac{4^2}{9^2}+\frac{y^2}{10^2}=1. \] Step 2. Compute the first term: \[ \frac{16}{81}+\frac{y^2}{100}=1. \] Step 3. Solve for \(\frac{y^2}{100}\) by subtracting \(\frac{16}{81}\) from both sides: \[ \frac{y^2}{100} = 1-\frac{16}{81}=\frac{81-16}{81}=\frac{65}{81}. \] Step 4. Multiply both sides by \(100\): \[ y^2 = 100\cdot\frac{65}{81}=\frac{6500}{81}. \] Step 5. Take the square root: \[ y = \sqrt{\frac{6500}{81