Answer:
Option D is correct
The maximum height of the ball was 2.5 seconds after it was thrown.
Step-by-step explanation:
let a and b are the zeros of the function f(x) then;
f(x)= k(x-a)(x-b)
where, k is the coefficient.
As per the statement:
A ball is thrown straight up into the air. The height, in feet, h(t), of the ball collected over t seconds is shown in the table.
From the given table:
At t = 0
h(0)= 0
At t = 5
h(5) = 0
⇒0 and 5 are the zeros the function h(t)
then by definition we have;
[tex]h(t) = k(t-0)(t-5)[/tex]
⇒[tex]h(t) = k(t)(t-5)[/tex] .....[1]
Now substitute any point from the table i.e, (2, 96) in [1] to find k;
[tex]96 = k(2)(2-5)[/tex]
[tex]96 = k(2)(-3)[/tex]
Simplify:
[tex]96 = -6k[/tex]
Divide both sides by -6 we have;
[tex]-16 = k[/tex]
or
k = -16
then, we have the equation for h(t) as:
[tex]h(t) = -16t(t-5)[/tex]
⇒[tex]h(t) = -16t^2+80t[/tex] ....[1]
Initial height of the ball:
h(0) = 0
To find the maximum height of the ball.
A quadratic equation [tex]y=ax^2+bx+c[/tex].....[2], the the axis of symmetry is given by:
[tex]x = \frac{-b}{2a}[/tex]
On comparing [1] and [2] we have;
a =-16 and b = 80 then;
[tex]t = \frac{-80}{2(-16)} = \frac{80}{32} = 2.5[/tex] sec
⇒the maximum height of the ball was 2.5 second.
Therefore, the statement which is true is: The maximum height of the ball was 2.5 seconds after it was thrown.
