The path of the football is an illustration of parabolas.
- The maximum height is: \(\mathbf{19.2525yd}\).
- The football is kicked to 51 yards
The function is given as:
\(\mathbf{y = -0.03x^2 + 1.53x}\)
(a) The maximum height
First, we differentiate the function with respect to x
\(\mathbf{y' = -0.06x + 1.53}\)
Set the above equation to 0
\(\mathbf{-0.06x + 1.53 = 0}\)
Collect like terms
\(\mathbf{-0.06x =- 1.53}\)
Divide both sides by -0.06
\(\mathbf{x =25.5}\)
Substitute 25.5 for x in \(\mathbf{y = -0.03x^2 + 1.53x}\)
\(\mathbf{y = -0.03 \times 25.5^2 + 1.52 \times 25.5}\)
\(\mathbf{y = 19.2525}\)
Hence, the maximum height is: \(\mathbf{19.2525yd}\)
(b) How far the football is kicked
Equate \(\mathbf{y = -0.03x^2 + 1.53x}\) to 0
\(\mathbf{-0.03x^2 + 1.53x = 0}\)
Now, we solve for x (i.e. the horizontal distance)
\(\mathbf{-0.03x^2 =- 1.53x }\)
Divide both sides by -x
\(\mathbf{0.03x =1.53 }\)
Divide both sides by 0.03
\(\mathbf{x =51 }\)
Hence, the football is kicked to 51 yards
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