Check the picture below.
[tex]~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&5\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&9\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-4.9t^2+5t+9\implies \stackrel{h(t)}{0}=-4.9t^2+5t+9 \implies 4.9t^2-5t-9=0[/tex]
since we don't get any neat integer values from it hmmm let's plug that in the quadratic formula to see what we get as h(t) = 0.
[tex]~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{4.9}t^2\stackrel{\stackrel{b}{\downarrow }}{-5}t\stackrel{\stackrel{c}{\downarrow }}{-9}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-5) \pm \sqrt { (-5)^2 -4(4.9)(-9)}}{2(4.9)}\implies t=\cfrac{5\pm\sqrt{25+176.4}}{9.8} \\\\\\ t=\cfrac{5\pm \sqrt{201.4}}{9.8}\implies t\approx \begin{cases} 1.96~~\checkmark\\ -0.94 \end{cases}[/tex]
notice, we do not use the negative value for "t", since the seconds cannot be less than 0.
