Question

The path of a skateboarder can be modeled by the equation (x – 3)2 = –8(y – 8), where distance is in feet and air resistance is insignificant. If the skateboarder makes a jump from a ramp at x = 0, what is her maximum height and how far is she horizontally from the end of the ramp when she lands?
Maximum height 3 feet; horizontal distance 16 feet
Maximum height 7 feet; horizontal distance 11 feet
Maximum height 8 feet; horizontal distance 11 feet
Maximum height 8 feet; horizontal distance 16 feet

Answer 1

Answer:

Maximum height 8 feet; horizontal distance 11 feet

Step-by-step explanation:

got it right

Answer 2

The maximum height and horizontal distance is required.

Maximum height 8 feet; horizontal distance 16 feet

The given path is $(x-3)^2=-8(y-8)$

If the skateboarder jumps at x = 0 she will land at y = 0. However there will be at horizontal displacement.

$(x-3)^2=-8(0-8)\\\Rightarrow (x-3)^2=8^2\\\Rightarrow x-3=\pm 8\\\Rightarrow x-3=8\\\Rightarrow x=11$

or

$x-3=-8\\\Rightarrow x=-5$

The distance between -5 and 11 is $11-(-5)=16$

Middle point would be $\dfrac{16}{2}=8$ from $-5$.

$-5+8=3$

So, $x=3$

The maximum height will be

$(3-3)^2=-8(y-8)\\\Rightarrow y-8=0\\\Rightarrow y=8$

The maximum height is 8 feet and the horizontal distance is 16 feet.

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