The following equation describes the height h of the ball after an amount of time of t seconds.
[tex]h(t)=-4.9t^2+2.94t+137.2[/tex]The ground represents the height zero, therefore, to find the corresponding time when the ball hits the ground we just have to solve the equation
[tex]-4.9t^2+2.94t+137.2=0[/tex]To solve this equation we can use the quadratic formula.
Our solutions are
[tex]\begin{gathered} t_{\pm}=\frac{-(2.94)\pm\sqrt{(2.94)^2-4(-4.9)(137.2)}}{2(-4.9)} \\ =\frac{-(2.94)\pm\sqrt{8.6436+2689.12}}{-9.8} \\ =\frac{2.94\pm\sqrt{2697.7636}}{9.8} \\ =\frac{2.94\pm51.94}{9.8} \end{gathered}[/tex]Our solution is only the positive root.
[tex]t=\frac{2.94+51.94}{9.8}=5.6[/tex]The object strikes the ground after 5.6 seconds.
