Answer:
A
[tex]h = -16t^2 + 192t[/tex]
B
Vertex=(6,576)
Step-by-step explanation:
The problem gives us the following data:
[tex]v_o=192\ ft/s,\ h_o=0[/tex]
A.
Thus the function is
[tex]h = -16t^2 + 192t[/tex]
The graph of h has the shape of an inverted parabola. Recall if the coefficient of the quadratic term is negative, the parabola is concave down, so it has a maximum value.
Part B
Let's take the function of h
[tex]h = -16t^2 + 192t[/tex]
Factoring by -16
[tex]\displaystyle h = -16(t^2 - 12t)[/tex]
Completing squares
[tex]\displaystyle h = -16(t^2 - 12t+36-36)[/tex]
[tex]\displaystyle h = -16(t^2 - 12t+36)+576[/tex]
Factoring
[tex]\displaystyle h = -16(t-6)^2+576[/tex]
Rearranging
[tex]\displaystyle h -576= -16(t-6)^2[/tex]
We can get the coordinates of the vertex from this standard form of the parabola.
Vertex=(6,576)
The maximum value means that at t=6 seconds, the firework will be 576 feet high and then it will start falling back to the ground.