We have the distance s in function of time expressed as:
[tex]s=8t+16t^2[/tex]
We then have to find t so that s = 80 feet.
We can then find it as:
[tex]\begin{gathered} s=80 \\ 8t+16t^2=80 \\ 8(t+2t^2)=80 \\ t+2t^2=\frac{80}{8} \\ 2t^2+t=10 \\ 2t^2+t-10=0 \end{gathered}[/tex]
We can calculate the solutions for this equation as:
[tex]\begin{gathered} t=\frac{-1\pm\sqrt{1^2-4(2)(-10)}}{2(2)} \\ t=\frac{-1\pm\sqrt{1+80}}{4} \\ t=\frac{-1\pm\sqrt{81}}{4} \\ t=\frac{-1\pm9}{4} \\ \Rightarrow t_1=\frac{-1-9}{4}=-\frac{10}{4}=-2.5 \\ \Rightarrow t_2=\frac{-1+9}{4}=\frac{8}{4}=2 \end{gathered}[/tex]
As time t has to be positive, the only valid solution in the context of this problem is t = 2 seconds.
Answer: It takes 2 seconds for the object to fall 80 feet.
