Answer:36.4 ft
Explanation:
Given
Span of Parabola [tex]L=100 ft[/tex]
Maximum height [tex]h=40 [/tex]
suppose Parabola is of type
[tex](x-x_0)^2=-4a(y-y_0)[/tex]
where [tex]x_0,y_0[/tex] is the center of parabola
[tex]x_0=0, y_0=40[/tex]
[tex]x^2=-4a(y-40)[/tex]
at [tex]y=0 [/tex]
[tex]x^2=-4a\times (-40)[/tex]
[tex]x^2=160a[/tex]
[tex]x=\pm \sqrt{160a}[/tex]
and it is given, [tex]2x=100[/tex]
[tex]x=50[/tex]
[tex]\sqrt{160a}=50[/tex]
[tex]a=15.625[/tex]
thus [tex]x^2=-4a(y-40)[/tex]
at [tex]x=15 [/tex]
[tex]15^2=-4\times 15.625(y-40)[/tex]
[tex]y=36.4 ft[/tex]