An object is launched upwardfrom 62.5 meters above ground level with aninitial velocity of 12 meters per second. Thegravitational pull of the earth is about 4.9meters per second squared. How long will theobject take to hit the ground?​

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.