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The height of a kicked football can be represented by the polynomial -167 +47t+ 3 O, where t is the time in seconds. Find the time in seconds of when the football hits the ground.

The height of a kicked football can be represented by the polynomial -167 +47t+ 3 O-example-1

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Polynomial of the height of a kicked football


-16t^2+47t+3=0

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:


\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

In this case you find the t: a=-16 b=47 c=3


\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=(-47\pm49)/(-32) \\ \\ t_1=(-47+49)/(-32)=-(2)/(32)=-(1)/(16) \\ \\ t_2=(-47-49)/(-32)=(-96)/(-32)=3 \end{gathered}

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
User Maksim Kondratyuk
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