Question 211525
<pre><font size = 4 color = "indigo"><b>
The football is kicked at point (0,0).
The highest point (vertex) is at (48,32)
The goalpost crossbar is at (90,10).

We begin by plotting the vertex, and drawing
a line for the goalpost.

{{{drawing(182,400,-1,100,.3,40,

graph(182,400,-1,100,.3,40),
line(90,0,90,10), locate(35,35,"(48,32)")
line(48+.3,32,48-.3,32), line(48,32+.3,48,32-.3), line(48-.3,32,48+.3,32),
line(48+.3,32+.3,48-.3,32-.3), line(48-.3,32-.3,48+.3,32+.3), line(48+.3,32-.3,48+.3,32-.3),locate(75,12,"(90,10)"),
line(48-.3,32+.3,48-.3,32+.3)

 )}}}

The equation of a parabola is

{{{y=a(x-h)^2+k}}} where the vertex is {{{"(h,k)"}}}

We substitute {{{"(h,k)"="(48,32)"}}}

{{{y=a(x-48)^2+32}}}

We can find {{{a}}} because the parabola goes
through the origin {{{"(0,0)"}}}

So we substitute {{{"(x,y)"="(0,0)"}}}

{{{y=a(x-48)^2+32}}}
{{{0=a(0-48)^2+32}}}
{{{0=a(-48)^2+32}}}
{{{0=a(2304)+32}}}
{{{0=2304a+32}}}
{{{-32=2304a}}}
{{{(-32/2304)=a}}}
{{{-1/72=a}}}

So the equation of the parabola is

{{{y=(-1/72)(x-48)^2+32}}}

We draw the parabola:

{{{drawing(182,400,-1,100,.3,40,
locate(75,12,"(90,10)"),
graph(182,400,-1,100,.3,40,(-1/72)(x-48)^2+32),
line(90,0,90,10), locate(35,34,"(48,32)")
line(48+.3,32,48-.3,32), line(48,32+.3,48,32-.3), line(48-.3,32,48+.3,32),
line(48+.3,32+.3,48-.3,32-.3), line(48-.3,32-.3,48+.3,32+.3), line(48+.3,32-.3,48+.3,32-.3),
line(48-.3,32+.3,48-.3,32+.3)

 )}}}

Oh,oh.  From the picture it appears that the
football is going to drop too low to go over
the goal post bar.  But we can't just go by the 
picture.  Let's calculate the height of the 
ball when {{{x=90}}} and find out for sure:

{{{y=(-1/72)(x-48)^2+32}}}
{{{y=(-1/72)(90-48)^2+32}}}
{{{y=(-1/72)(42)^2+32}}}
{{{y=(-1/72)(1764)+32}}}
{{{y=(-1764/72)+32}}}
{{{y=-24.5+32}}}
{{{y=7.5}}}

No, the field goal will not be made, because
the ball will be too low, and will go under
the goal post, since 7.5 feet is less
than 10 feet.

Edwin</pre>