if the parkway is 60 feet wide, the minimum clearance is 15 feet and the parabolic arch is 25 feet high, how long is the arch?

Answer 1

If the clearance is 15 ft at the edges of the pavement, then we have
y = 25-ax^2
y(30) = 15
so now we know that
25 - 900a = 15
a = 1/90
now, knowing that
y = 25 - 1/90 x^2, we have
y=0 at x = 15√10 ≈ 47.43
so the arch is 94.86 ft wide at its base

I assume you are not really after the actual length of the curve, since that requires calculus.