If a stone is dropped from a height of 256 ft, then its height (in feet) above the ground is given by the function h(t)equalsnegative 16 t squared plus 256 where t is time (in seconds). to get an idea of how fast the stone is traveling when it hits the ground, find the average rate of change of the height on each of the time intervals [0,4], [1,4], [3.9,4], [3.99,4], and [3.999,4].

For [0,4] : -64

For [1,4] : -80

For [3.9,4] : -126.4

For [3.99,4] : -127.84

For [3.999,4] :  -127.984

Step-by-step explanation:

First we establish that we have the next function:

                                        h(t) = -16t² + 256

Now, to find the average rate of change we have to define the change in the output of the function and the change in the input of the function:

  Average rate of change = Change in the output / Change in the input

Which in this case will be represented by:

                        Average rate of change = h(t2) - h(t1) / t2 - t1

So now we resolve the function for each time interval and calculare the average rate of change:

For [0,4] :

h(0) = -16(0)² + 256 = 256

h(4) = -16(4)² + 256 = 0

Average rate of change = [0 - 256] / [4 - 0] = -64

For [1,4] :

h(1) = -16(1)² + 256 = 240

h(4) = -16(4)² + 256 = 0

Average rate of change = [0 - 240] / [4 - 1] = -80

For [3.9,4] :

h(3.9) = -16(3.9)² + 256 = 12.64

h(4) = -16(4)² + 256 = 0

Average rate of change = [0 - 12.64] / [4 - 3.9] = -126.4

For [3.99,4] :

h(3.99) = -16(3.99)² + 256 = 1.2784

h(4) = -16(4)² + 256 = 0

Average rate of change = [0 - 1.2784] / [4 - 3.99] = -127.84

For [3.999,4] :

h(3.999) = -16(3.999)² + 256 = 0.127984

h(4) = -16(4)² + 256 = 0

Average rate of change = [0 - 0.127984] / [4 - 3.999] = -127.984