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

Answer:

2 seconds

Step-by-step 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.

Step-by-step 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

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

$x = \frac{1 +- \sqrt{1 + 48 } }{4}$

$x = \frac{1 +- \sqrt{49 } }{4}$

$x = \frac{1 +- 7 }{4}$

x = 2, -3/2

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