check the picture below, the height there uses feet, but nevermind that, we can simply let it be meters, like in this case.
[tex]\bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in meters}
\\\\
h(t) = -4.9t^2+v_ot+h_o
\end{array}
\quad
\begin{cases}
v_o=\stackrel{25}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{3}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-4.9t^2+25t+3[/tex]
how hight does it go? well, that'd be the y-coordinate of the vertex.
how long does it take? well, that'd be the x-coordinate of the vertex.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{lcccl}
h(t) = & -4.9t^2& +25t& +3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(-\cfrac{25}{2(-4.9)}~~,~~3-\cfrac{25^2}{4(-4.9)} \right)\implies \left( \cfrac{25}{9.8}~~,~~3-\cfrac{625}{-19.6} \right)
\\\\\\
\left( \cfrac{25}{9.8}~~,~~3+\cfrac{625}{19.6} \right)\implies \left( \cfrac{25}{9.8}~~,~~\cfrac{3419}{98} \right)[/tex]
