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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 $y$ (in feet) of a second hockey puck is modeled by $y=x\left(0.15-0.001x\right)$ , where $x$ is the horizontal distance (in feet). Compare the distances traveled by the hockey pucks before hitting the ground.

Respuesta :

Answer:

Second puck travels farther

Step-by-step explanation:

Maximum height of first puck = 3 feet

The height of a second hockey puck is modeled by:

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

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

To find maximum height of second puck

[tex]y'=0.15-0.002x[/tex]

Equate the derivative equals to 0

0.15-0.002x=0

[tex]\frac{0.15}{0.002}=x[/tex]

75 = x

At x = 75

[tex]y=0.15(75)-0.001(75)^2=5.625[/tex]

So, The maximum height of second puck is greater than  first puck

So, Second puck travels farther

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