Question

A hockey player strikes a hockey puck. The height of the puck increases until it reaches and 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(0.15-0.001x), where x is the horizontal distance (in feet). Compare the distances traveled by the hockey pucks before hitting the ground.

Answer 1

Distance traveled by hockey puck of second player is 40 feet more than that of first player.

Explanation:

Maximum height of 3 feet is attained by the hockey puck of first player is when the puck is 55 feet away from the player.

It is at the mid point of the total horizontal distance to be traveled by the hockey puck.

So, the horizontal distance traveled by hockey puck of first player = 55 + 55 = 110 feet

Second player:

`y=x(0.15-0.001x)`

where
`y` is the height of a second hockey puck

`x` is the horizontal distance.

When the puck hits the ground,
`y=0`

Now, let us find the value of
`x`.

`\Rightarrow 0=x(0.15-0.001x)`

At
`x=0` feet or
`x=150` feet,
`y=0`

Therefore, the distance traveled by second hockey puck is 150 feet.

So, the answer is:

Distance traveled by hockey puck of second player is 40 feet more than that of first player.