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A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet away from the player. The height $y$y​ (in feet) of a second hockey puck is modeled by $y=x\left(0.15-0.001x\right)$y=x(0.15−0.001x)​ , where x$x$x​ is the horizontal distance (in feet). Compare the distances traveled by the hockey pucks before hitting the ground.

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Answer:

The maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels further  

Also the distance of the maximum height of first pluck from the player is less than the distance of the second pluck from the player hence the second pluck travels further

Step-by-step explanation:

From the question we are told that

  The maximum height of the first pluck is  [tex]h_1 = 3 \ ft[/tex]

   The height of the second height is mathematically represented as

              [tex]$y=x\left(0.15-0.001x\right)[/tex]

=>           [tex]y=0.15x -0.001x^2[/tex]

Generally at maximum height  [tex]y' = 0[/tex]

So

      [tex]y'=0.15 -0.002 x = 0[/tex]

=>    [tex]x =  75 \  ft [/tex]

Here 75 ft is the horizontal distance the second pluck traveled at maximum height

   So the maximum height of the second pluck is mathematically represented as

       [tex]y=0.15(75) -0.001(75)^2[/tex]

=>     [tex]y=  5.625 \  ft [/tex]  

So comparing the maximum height  of the first and the second pluck we see that the maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels father

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