lets assume that the ground level corresponds to y=0, then is we

solve 16-0.24 y^2=0, we will find the two points that the arch touches

ground and the width will be the difference between them, so:

16-0.24y^2=0,

or:

0.25 y^2=16

taking square roots:

(1/2) y= +/- 4

y=-8 or +8.

So the width of the arch is 8-(-8)=16.

RonL