We got to use the given formula:
[tex]d=v\cdot t+16t^2[/tex]The distance, d, given is 320 ft and the initial velocity, v, 16 ft/s. We want the time, t. So:
[tex]\begin{gathered} d=v\cdot t+16t^2 \\ 320=16t+16t^2 \\ 16t^2+16t-320=0 \\ \frac{16t^2}{16}+\frac{16t}{16}-\frac{320}{16}=\frac{0}{16} \\ t^2+t-20=0 \end{gathered}[/tex]Now, we have a quadratic equation, so we can use Bhaskara formula:
[tex]\begin{gathered} t=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-20)}}{2\cdot1}=\frac{-1\pm\sqrt[]{1+80}}{2}=\frac{-1\pm\sqrt[]{81}}{2}=\frac{-1\pm9}{2} \\ t_1=\frac{-1-9}{2}=-\frac{10}{2}=-5 \\ t_2=\frac{-1+9}{2}=\frac{8}{2}=4 \end{gathered}[/tex]Because we can't have a negative time, we consider only the second one, which it t = 4s.
