Question

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​].

Answer 1

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

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