Polynomial of the height of a kicked football
[tex]-16t^2+47t+3=0[/tex]As the equation above represents the height, to find the time in second of when the footbal hits the groud you equal the equation to 0 (0 is the height of the ground) and solve for t:
To solve an equation as given (quadratic equiation) you use the next formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In this case you find the t: a=-16 b=47 c=3
[tex]\begin{gathered} t=\frac{-47\pm\sqrt[]{47^2-4(-16)(3)}}{2(-16)} \\ \\ t=\frac{-47\pm\sqrt[]{2209-(-192)}}{-32} \\ \\ t=\frac{-47\pm\sqrt[]{2401}}{-32} \\ \\ t=\frac{-47\pm49}{-32} \\ \\ t_1=\frac{-47+49}{-32}=-\frac{2}{32}=-\frac{1}{16} \\ \\ t_2=\frac{-47-49}{-32}=\frac{-96}{-32}=3 \end{gathered}[/tex]As you get two solutions for t, and the time cannot be negative, the correct solution in this situation is t=3
Then, the time the football hits the groud is 3 seconds