check the picture below.
now, keeping in mind that gravity will pull the projectile down, and that'd be 9.8 m/s², that gets halfed for the initial velocity equation, so 9.8/2 or 4.9 m/s².
[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{98}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{\textit{from the ground thus }0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-4.9t^2+98t+0[/tex]
how high will it go? well, notice in the picture, it'll go as high as its vertex,
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -4.9}}t^2&{{ +98}}t&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left(\stackrel{\textit{how long it took}}{-\cfrac{98^2}{2(-4.9)}}~~,~~\stackrel{\textit{how high it went}}{0-\cfrac{98^2}{4(-4.9)}} \right)
\\\\\\
0-\cfrac{9604}{-19.6}\implies 490[/tex]
when will it return to the ground? well, notice the picture, when y = 0.
[tex]\bf \stackrel{h(t)}{0}=-4.9t^2+98t+0\implies 4.9t^2-98t=0\implies t(4.9t-98)=0\\\\
-------------------------------\\\\
t=0\\\\
-------------------------------\\\\
4.9t-98=0\implies 4.9t=98\implies t=\cfrac{98}{4.9}[/tex]
so after "t" seconds it comes back down.
notice is 0 at two points, at t=0 or 0 seconds, because it just took off the ground, and again the second time, when it comes back down.
