Question

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards, 30 yards away from the player. A second kick is modeled by f(x)=−0.032x(x−50), where f is the height (in yards) and x is the horizontal distance (in yards). Compare the distances that the footballs travel before hitting the ground.

Answer 1

The second ball traveled more distance, horizontally and vertically.

Step-by-step explanation:

The given function is

`f(x)=-0.032x(x-50)`

To find the distances that the second football travels, we need to find the vertex of its movement, because it's movement has a parabola form. The quadratic expression is

`f(x)=-0.032x^(2) +1.6x`

Where
`a=-0.032` and
`b=1.6`

The vertex has coordinates of
`(h,k)`, where

`h=-(b)/(2a)`

Replacing values, we have

`h=-(1.6)/(2(-0.032))=25`

Then,
`k=f(h)`

`k=f(25)=-0.032(25)^(2) +1.6(25)\\k=-20+40=20`

Which means the maximum height of the second football is 20 yards. That means it travels 40 yards vertically.

Now, its horizontal distance can be found when
`f(x)=0`

`0=-0.032x^(2) +1.6x\\0=x(-0.032x+1.6)\\x_(1)=0\\ -0.032x_(2) +16=0\\x_(2)=(-16)/(-0.032)\\ x_(2)=500`

So, its horizontal distance is 500 yards.

Comparing the distances between the footballs.

Ball 1

Horizontal distance of 30 yards.

Vertical distance of 30 yards.

Ball 2

Horizontal distance of 500 yards.

Vertical distance of 40 yards.

If we find their difference, it would be

Horizontal: 500 - 30 = 470 yards.

Vertical: 40 - 30 = 10 yards.

Therefore, the second ball traveled more distance, horizontally and vertically.

Answer 2

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Step-by-step explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

`Y = 1.6 X - 0.032x^2`

we know that maximum height occurs is given as

`x = -(b)/(2a)`

`y =- (1.6)/(2(-0.032)) = 25`

and maximum height is

`y = 1.6* 25 - 0.032* 25^2`

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards