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:
[tex]x = \frac{-b +- \sqrt{b^{2} - 4ac } }{2a}[/tex]
here,
a = 2
b = -1
c = -6
Putting these values in formula, we get
[tex]x = \frac{-(-1) +- \sqrt{-1^{2} - 4.2.-6 } }{2.2}[/tex]
[tex]x = \frac{1 +- \sqrt{1 + 48 } }{4}[/tex]
[tex]x = \frac{1 +- \sqrt{49 } }{4}[/tex]
[tex]x = \frac{1 +- 7 }{4}[/tex]
x = 2, -3/2
As the time cannot be negative. Therefore, the ball will hit the ground at 2 seconds.
Answer:
2 seconds
Step-by-step explanation:
The equation that models the height of the soccer ball is
[tex]h(x) = - 2 {x}^{2} + x + 6[/tex]
When the soccer ball hit the ground, then the height is zero.
This implies that:
[tex] - 2 {x}^{2} + x + 6 = 0[/tex]
We split the middle term to get:
[tex] - 2 {x}^{2} + 4x - 3x + 6 = 0[/tex]
[tex] - 2x(x - 2) - 3(x - 2) = 0[/tex]
We factor to get:
[tex](x - 2)( - 2x - 3) = 0[/tex]
[tex]x = 2 \: or \: x = - 1.5[/tex]
Since time must be positive, we have x=2
