Check the picture below. The picture is using feet, but is pretty much the same trajectory for meters.
so the object hits the ground when y = 0, thus
[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{12}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{62.5}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-4.9t^2+12t+62.5\implies \stackrel{\textit{hitting the ground}}{\stackrel{h(t)}{0}=-4.9t^2+12t+62.5}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-4.9}t^2\stackrel{\stackrel{b}{\downarrow }}{+12}t\stackrel{\stackrel{c}{\downarrow }}{+62.5} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-12\pm\sqrt{12^2-4(-4.9)(62.5)}}{2(-4.9)}\implies t=\cfrac{-12\pm\sqrt{144+1225}}{-9.8} \\\\\\ t=\cfrac{12\mp\sqrt{1369}}{9.8}\implies t=\cfrac{12\mp 37}{9.8}\implies t= \begin{cases} \stackrel{\approx}{-2.55}\\ 5 \end{cases}[/tex]
we can't use -2.55, since it's seconds and can't be negative, so t = 5 seconds later.
