Question 1160972
<pre>
Place the vertex at (0,40) and the ends of the bridge at (±90,0).

{{{drawing(800, 240, -100,100, -10,50,
line(10,0,10,40),
graph(800, 240, -100,100, -10,50, 40 - (2(x)^2)/405) )}}}

The equation is

{{{(x-h)^2=4p(y-k)}}}

Where the vertex (h,k) = (0,40)

{{{(x-0)^2=4p(y-40)}}}

{{{x^2=4p(y-40)}}}

Substituting the point (90,0)

{{{90^2=4p(-40)}}}

{{{8100=-160p}}}

{{{-405/8=p}}}

So the equation is now

{{{x^2=4(-405/8)(y-40)}}}

{{{x^2=(-405/2)(y-40)}}}

So when x=10

{{{10^2=(-405/2)(y-40)}}}

{{{100=(-405/2)(y-40)}}}

{{{200=-405(y-40)}}}

{{{200=-405y+16200}}}

{{{405y=16200}}}

{{{y=16200/405}}}

{{{y=40}}}

Yes a 39 foot tall sailboat can fit under there with 1 foot to spare.

Edwin</pre>