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