Answer:
The ball touches the ground in a time t = 3.02 seconds.
Step-by-step explanation:
We have the height equation as a function of time:
h = 182−12t-16t² Equation (1)
The height is equal to zero when the ball touches the ground.
Then we replace h = 0 in equation (1) to calculate the time in which the ball touches the ground:
0 = 182−12t-16t²
16t²+ 12t -182 = 0: Quadratic equation
We solve the quadratic equation to calculate t:
[tex]t_{1} =\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]t_{1} =\frac{-12+\sqrt{-12^{2}-4*16*-182 } }{2*16}[/tex]
[tex]t_{2} =\frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]t_{2} =\frac{-12-\sqrt{-12^{2}-4*16*-182 } }{2*16}[/tex]
a = 16, b = 12, c = -182
[tex]t_{1} =3.02 s[/tex]
[tex]t_{2} =-3.76 s[/tex]
We take only the positive value for time because negative time does not exist,then,the ball touches the ground in a time t = 3.02 seconds.
