Question 1005396
The parabola needed, drawn over a conveniently placed set of axes, with all measures in feet, looks like this:
{{{drawing(400,400,-70,70,-20,80,
rectangle(-40,0,40,20),
locate(18,20,40),locate(-22,20,40),
locate(34,12,20),
graph(400,400,-70,70,-20,80,20,-x^2/45+500/9)
)}}} The equation will be {{{y=f(x)=ax^2+k}}} such that {{{system(f(40)=20,f(50)=0)}}} .
{{{system(y=f(x)=ax^2+k,f(40)=20,f(50)=0)}}}--->{{{system(y=f(x)=ax^2+k,a40^2+k=20,a50^2+k=20)}}}--->{{{system(y=f(x)=ax^2+k,1600a+k=20,2500a+k=0)}}}--->{{{system(y=f(x)=-x^2/45+500/9,a=-1/45,k=500/9)}}}
The height at the vertex (in feet) is {{{f(0)=highlight(500/9="55.55 ( rounded )")}}} .