Answer:
Ok, we know that the baseball will have an equation like:
h(x) = a*x^2 + b*x + c
Where x represents time, and h(x) represents height.
Let's construct it:
The acceleration is:
a(t) = a
For the velocity, we can integrate over time and get:
v(x) = a*x + v0
where v0 is the initial vertical velocity.
and for the position (or the height) we can integrate again:
h(x) = a*x^2 + v0*x + h0
Where h0 is the initial position.
Then our equation is:
h(x) = a*x^2 + v0*x + h0.
Now let's look at the table:
when x = 0s, h(0s) = 6ft
Then:
h(0s) = a*0s^2 + v0*0s + h0 = 6ft
h0 = 6ft.
We also can see that:
h(2s) = h(4s)
Knowing that the quadratic function has symmetry around a point, we can conclude that the point of symmetry is right in between of 2s and 4s, which is at x = 3s.
Now, for a standard quadratic equation:
a*x^2 + b*x + c
The symmetry line is at:
x = -b/2a
In this case:
b = v0
a = a
then we have:
3s = -v0/(2*a)
v0 = -3s*(2a)
Now we have all the data that we need for our equation, we can write it as:
h(x) = a*x^2 - 6s*a*x + 6ft
Now we have only one variable left, we know that at x = 2s, h(2s) = 22ft
then:
h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft
22ft - 6ft = 16ft = a*(4s^2 - 12s^2) = a*(-8s^2)
16ft/(-8s^2) = a = -2ft/s^2
Then our equation is:
h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft
b) The height after 5 seconds is given by:
h(5s) = (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft
