Complete the square:
[tex]h(t)=-16t^2+36t+9=-16\left(t^2-\dfrac94t\right)+9=-16\left(\left(t-\dfrac98\right)^2-\dfrac{81}{64}\right)+9[/tex]
[tex]h(t)=-16\left(t-\dfrac98\right)^2+\dfrac{117}4[/tex]
That is, the graph of [tex]h(t)[/tex] is a parabola with vertex at [tex](t,h(t))=\left(\dfrac98,\dfrac{117}4\right)[/tex], meaning the ball reaches its maximum height of [tex]\dfrac{117}4=29.25\,\mathrm{ft}[/tex] after [tex]t=\dfrac98=1.125\,\mathrm s[/tex] of being thrown.
