We know the distance the object will travel is equal to the height of the cliff. Therefore,
[tex]d(t)=4.9t^2+12t^{}\rightarrow420=4.9t^2+12t[/tex]We then clear t. Notice we get a quadratic equation
[tex]\begin{gathered} 4.9t^2+12t=420\rightarrow4.9t^2+12t-420=0 \\ \text{Solve using the general formula} \\ x=\frac{-12\pm\sqrt[]{12^2-4(4.9)(-420)}}{2(4.9)} \\ \rightarrow x=\frac{-12\pm\sqrt[]{8376}}{9.8} \\ \rightarrow x_1=8.11s \\ \rightarrow x_2=-10.56s \end{gathered}[/tex]We discard the negative solution, because negative time doesn't exist.
Therefore, it takes the object 8.11s to hit the ground.
