Question

A soccer ball was kicked in the air and follows the path h(x)=−2x2+1x+6, where x is the time in seconds and h is the height of the soccer ball. At what time will the soccer ball hit the ground

Answer 1

2 seconds

Explanation:

The equation that models the height of the soccer ball is

`h(x) = - 2 {x}^(2) + x + 6`

When the soccer ball hit the ground, then the height is zero.

This implies that:

`- 2 {x}^(2) + x + 6 = 0`

We split the middle term to get:

`- 2 {x}^(2) + 4x - 3x + 6 = 0`

`- 2x(x - 2) - 3(x - 2) = 0`

We factor to get:

`(x - 2)( - 2x - 3) = 0`

`x = 2 \: or \: x = - 1.5`

Since time must be positive, we have x=2

Answer 2

The ball will hit the ground at 2 seconds.

Explanation:

Given that,

The path of the ball = h(x)=−2x2+1x+6

Here,

x is the time while h is the height of ball.

When the ball will hit the ground, the height will become zero. Therefore,

h(x)=−2x2+1x+6

0 =−2x2+1x+6

or

2x2 -1x - 6 = 0

This is a quadratic equation, hence by applying quadratic equation formula:

`x = \frac{-b +- \sqrt{b^(2) - 4ac } }{2a}`

here,

a = 2

b = -1

c = -6

Putting these values in formula, we get

`x = \frac{-(-1) +- \sqrt{-1^(2) - 4.2.-6 } }{2.2}`

`x = (1 +- √(1 + 48 ) )/(4)`

`x = (1 +- √(49 ) )/(4)`

`x = (1 +- 7 )/(4)`

x = 2, -3/2

As the time cannot be negative. Therefore, the ball will hit the ground at 2 seconds.