Answer:
16-5
Step-by-step explanation:
Step-by-step explanation:
The soccer ball was kicked in the air and follows the path as :
![h(x)=-2x^2+x+6](https://tex.z-dn.net/?f=h%28x%29%3D-2x%5E2%2Bx%2B6)
Here, x is the time in seconds and h is the height of the soccer ball.
We need to find the time at which the soccer ball hit the ground. It means that its height at a function of time becomes 0. So,
h(x) = 0
![-2x^2+x+6=0\\\\2x^2-x-6=0](https://tex.z-dn.net/?f=-2x%5E2%2Bx%2B6%3D0%5C%5C%5C%5C2x%5E2-x-6%3D0)
The above equation is a quadratic equation. It can be calculated as :
![x= \dfrac{-b \pm \sqrt{b^{2} - 4ac } }{2a}\\\\x= \dfrac{-b + \sqrt{b^{2} - 4ac } }{2a},x= \dfrac{-b - \sqrt{b^{2} - 4ac } }{2}\\\\x= \dfrac{-(-1) + \sqrt{(-1)^{2} - 4(2)(-6) } }{2(2)},x= \dfrac{-(-1) - \sqrt{(-1)^{2} - 4(2)(-6) } }{2(2)}\\\\x=2,-1.5](https://tex.z-dn.net/?f=x%3D%20%5Cdfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D%20-%204ac%20%7D%20%20%20%7D%7B2a%7D%5C%5C%5C%5Cx%3D%20%5Cdfrac%7B-b%20%2B%20%5Csqrt%7Bb%5E%7B2%7D%20-%204ac%20%7D%20%20%20%7D%7B2a%7D%2Cx%3D%20%5Cdfrac%7B-b%20-%20%5Csqrt%7Bb%5E%7B2%7D%20-%204ac%20%7D%20%20%20%7D%7B2%7D%5C%5C%5C%5Cx%3D%20%5Cdfrac%7B-%28-1%29%20%2B%20%5Csqrt%7B%28-1%29%5E%7B2%7D%20-%204%282%29%28-6%29%20%7D%20%20%20%7D%7B2%282%29%7D%2Cx%3D%20%5Cdfrac%7B-%28-1%29%20-%20%5Csqrt%7B%28-1%29%5E%7B2%7D%20-%204%282%29%28-6%29%20%7D%20%20%20%7D%7B2%282%29%7D%5C%5C%5C%5Cx%3D2%2C-1.5)
So, at 2 seconds the ball will hit the ground.