Question

A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds.

Answer 1

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