check the picture below.
so, since the ball kick is a parabola opening downwards, like you see in the picture, the axis of symmetry will come from the x-coordinate of the vertex, so let's find it.
[tex]\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{\textit{0, since it was on the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+64t\implies h(t)=-16t^2+64t+0 \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{64}{2(-16)}~,\qquad \qquad \right)\implies (2~,~\qquad )~\hfill \stackrel{\textit{axis of symmetry}}{x=2} \\\\\\ ~\hspace{34em}[/tex]
