Question

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.

Answer 1

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:

$y=x\left(0.15-0.001x\right)$

$y=0.15x-0.001x^2$

To find maximum height of second puck

$y'=0.15-0.002x$

Equate the derivative equals to 0

0.15-0.002x=0

$\frac{0.15}{0.002}=x$

75 = x

At x = 75

$y=0.15(75)-0.001(75)^2=5.625$

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

So, Second puck travels farther