Answer:
A. the volleyball reached it's maximum height of 3 sec
B. The maximum height of the vollybal was 23 ft
E. The vball was served from a height of 5 ft
Step-by-step explanation:
h(t) = -2 (t-3)^2 +23
Given equation is in the form of [tex]f(x)= a(x-h)^2 + k[/tex]
(h,k) is the vertex
Now we compare f(x) with h(t)
h(t) = -2 (t-3)^2 +23
h = 3 and k = 23
Vertex is (3,23)
h=3 . this means the volleyball reaches its maximum height in 3 seconds
k = 23. this means the volleyball reaches the maximum height of 23 ft
When ball reaches the ground the height becomes 0. so plug in 0 for h(t) and solve for t
0= -2 (t-3)^2 +23
Subtract 23 on both sides
-23 = -2(t-3)^2
Divide both sides by -2
[tex]\frac{-23}{-2} = (t-3)^2[/tex]
Take square root on both sides
[tex]+-\sqrt{\frac{23}{2}}= t-3[/tex]
Add 3 on both sides
[tex]+-\sqrt{\frac{23}{2}}+3= t[/tex]
We will get two value for t
t=-0.39 and t= 6.39
So option C is not correct
Given h(t) is a quadratic function not exponential
To find initial height we plug in 0 for x and find out h(0)
h(0) = -2 (0-3)^2 +23 = -2(-3)^2 + 23= -18+ 23= 5
The volleyball was served from a height of 5 ft
