Respuesta :
Answer: The ball hits the ground at 5 s
Step-by-step explanation:
The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:
[tex]f=30 t- 6t^{2}[/tex] (1)
Where [tex]f[/tex] models the height of the ball in meters and [tex]t[/tex] the time.
Now, let's find the time [tex]t[/tex] when the ball Sara kicked hits the ground (this is when [tex]f=0 m[/tex]):
[tex]0=30 t- 6t^{2}[/tex] (2)
Rearranging the equation:
[tex]6t^{2}-30 t=0[/tex] (3)
Dividing both sides of the equation by [tex]6[/tex]:
[tex]t^{2}-5 t=0[/tex] (4)
This quadratic equation can be written in the form [tex]at^{2}+bt+c=0[/tex], and can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (5)
Where:
[tex]a=1[/tex]
[tex]b=-5[/tex]
[tex]c=0[/tex]
Substituting the known values:
[tex]t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(0)}}{2(1)}[/tex] (6)
Solving we have the following result:
[tex]t=5 s[/tex] This means the ball hit the ground 5 seconds after it was kicked by Sara.
Answer: Sarah kicked the ball from a height of about 1\text{ m}1 m1, start text, space, m, end text.
At its highest point, the ball was about 14\text{ m}14 m14, start text, space, m, end text above the ground.
Step-by-step explanation:
