Final answer:
To find the horizontal distance from the center of a bridge arch in the shape of half an ellipse where the height is 14.8 feet, we use the equation of an ellipse and solve for the variable x. Given the bridge is 150 feet wide and 32.4 feet high, the horizontal distance where the height is 14.8 feet is approximately 30.5 feet.
Step-by-step explanation:
The bridge arch described in the question is shaped like half an ellipse. To find the horizontal distance from the center where the height is equal to 14.8 feet, we will use the equation of an ellipse centered at the origin (x^2/a^2) + (y^2/b^2) = 1, where a is the semi-major axis and b is the semi-minor axis.
In this case, since the bridge is 150 feet wide, the entire length represents the major axis of the ellipse, so the semi-major axis a is 150/2 = 75 feet. The full height of 32.4 feet represents the minor axis, so the semi-minor axis b is 32.4/2 = 16.2 feet. Therefore, the equation of the ellipse is (x^2/75^2) + (y^2/16.2^2) = 1.
To find the horizontal distance x where the height y is equal to 14.8 feet, we substitute y into the equation and solve for x:
(x^2/75^2) + (14.8^2/16.2^2) = 1
(x^2/75^2) = 1 - (14.8^2/16.2^2)
x^2 = 75^2 * (1 - (14.8^2/16.2^2))
x = ±75 * sqrt(1 - (14.8^2/16.2^2))
We only consider the positive value of x since we are looking for a horizontal distance, which is always positive:
x ≈ ±75 * sqrt(1 - (219.04/262.44))
x ≈ ±75 * sqrt(1 - 0.8347)
x ≈ ±75 * sqrt(0.1653)
x ≈ ±75 * 0.4066
x ≈ ±30.5 feet
So, the horizontal distance from the center of the arch where the height is 14.8 feet is approximately 30.5 feet.