chech the picture below.
if the person kicked it from the ground, that means its initial height is 0.
it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.
[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{8}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+8t+0[/tex]
now let's find the y-coordinate of its vertex
[tex]\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(\qquad ,~~0-\cfrac{8^2}{4(-16)} \right)\implies (\quad ,~0+1)\implies (\quad ,~1)[/tex]
when will it hit the ground?
[tex]\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0
\\\\\\
8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}
\\\\\\
2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it when it came back down}}{t=\cfrac{1}{2}}[/tex]
