Respuesta :
Answer:
C: maximum height=62.25 feet; time=3.85 seconds
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
In this question:
The height of the ball, after t seconds, is given by the following equation:
[tex]h(t) = -16t^2 + 60t + 6[/tex]
Which is a quadratic equation with [tex]a = -16, b = 60, c = 6[/tex]
Maximum height:
Since a < 0, we can find the maximum value of the function. We have that:
[tex]\Delta = 60^{2} - 4(-16)(6) = 3984[/tex]
[tex]y_{v} = -\frac{3984}{4(-16)} = 62.25[/tex]
The maximum height is of 62.25 feet.
Seconds to reach the ground:
[tex]x_{1} = \frac{-60 + \sqrt{3984}}{2*(-16)} = -0.1[/tex]
[tex]x_{2} = \frac{-60 - \sqrt{3984}}{2*(-16)} = 3.85[/tex]
Since time is a positive measure, 3.85 seconds.
The correct answer is given by option C.
