Answer :
Answer:
The maximum height that the ball reaches is of 3.96 meters.
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].
In this question:
The height after t seconds is given by:
[tex]h(t) = -51t^2 + 20t + 2[/tex]
Which is a quadratic equation with [tex]a = -51, b = 20, c = 2[/tex]
What is the maximum height that the ball reaches?
This is [tex]h_v[/tex].
[tex]\Delta = b^2-4ac = 20^2 - 4(-51)(2) = 808[/tex]
[tex]y_{v} = -\frac{\Delta}{4a} = -\frac{808}{4(-51)} = 3.96[/tex]
The maximum height that the ball reaches is of 3.96 meters.