Answer:
Maximum height 8 feet; horizontal distance 11 feet
Step-by-step explanation:
got it right
The maximum height and horizontal distance is required.
Maximum height 8 feet; horizontal distance 16 feet
The given path is [tex](x-3)^2=-8(y-8)[/tex]
If the skateboarder jumps at x = 0 she will land at y = 0. However there will be at horizontal displacement.
[tex](x-3)^2=-8(0-8)\\\Rightarrow (x-3)^2=8^2\\\Rightarrow x-3=\pm 8\\\Rightarrow x-3=8\\\Rightarrow x=11[/tex]
or
[tex]x-3=-8\\\Rightarrow x=-5[/tex]
The distance between -5 and 11 is [tex]11-(-5)=16[/tex]
Middle point would be [tex]\dfrac{16}{2}=8[/tex] from [tex]-5[/tex].
[tex]-5+8=3[/tex]
So, [tex]x=3[/tex]
The maximum height will be
[tex](3-3)^2=-8(y-8)\\\Rightarrow y-8=0\\\Rightarrow y=8[/tex]
The maximum height is 8 feet and the horizontal distance is 16 feet.
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