Question 1087820
NOTE: I am going to round the answer to 2 decimal places, because the rounding required was not clearly stated.
 
Here is a cross-section of that tunnel
(with conveniently placed x- and y-axes):
{{{drawing(480,250,-24,24,-5,20,
red(arrow(12,0,24,0)),red(arrow(-12,0,-24,0)),
red(arrow(0,7.5,0,20)),red(arrow(0,7.5,0,-5)),
line(-18,0,18,0),arc(0,0,36,30,180,360),
locate(8,1.5,18ft),locate(-10,1.5,18ft),
locate(0.03,8,15ft),locate(23.7,1.7,red(x)),
locate(0.8,20.5,red(y))
)}}} . The ellipse equation is {{{x^2/18^2+y^2/15^2=1}}}
Here is the tallest 12-foot wide truck places exactly in the middle of that tunnel:
{{{drawing(480,250,-24,24,-5,20,
red(arrow(12,0,24,0)),red(arrow(-12,0,-24,0)),
red(arrow(0,7.5,0,20)),red(arrow(0,7.5,0,-5)),
line(-18,0,18,0),arc(0,0,36,30,180,360),
locate(8,1.5,18ft),locate(-10,1.5,18ft),
locate(2,5,6ft),locate(-4,5,6ft),
blue(arrow(6,7.07,6,0)),blue(arrow(6,7.07,6,14.14)),
locate(6.2,8,blue(h)),locate(23.7,1.7,red(x)),
locate(0.8,20.5,red(y)),rectangle(6,14.14,-6,3.5),
rectangle(4.3,0,3.3,3.5),rectangle(4.5,0,5.5,3.5),
rectangle(-4.3,0,-3.3,3.5),rectangle(-4.5,0,-5.5,3.5)
)}}} .
For that ruck to fit inside the tunnel {{{h}}} must be
at most the ellipse value for {{{y}}} when {{{x=6}}} .
{{{6^2/18^2+y^2/15^2=1}}}
{{{(6/18)^2+y^2/15^2=1}}}
{{{(1/3)^2+y^2/15^2=1}}}
{{{1/9+y^2/15^2=1}}}
{{{y^2/15^2=1-1/9}}}
{{{y^2/15^2=8/9}}}
{{{y^2=15^2*8/3^2}}}
{{{y^2=3^2*5^2*2^2*2/3^2}}}
{{{y^2=5^2*2^2*2}}}
{{{y^2=(5*2)^2*2}}}
{{{y^2=10^2*2}}}
{{{y=10sqrt(2)=about 14.14}}} .
So, if a truck is 12ft wide, for it to be able to fit through the tunnel,
the top of the truck must be at most {{{highlight(14.14ft)}}} high.