Question

3) The height of an object that is thrown straight up from a height of 6 feet

above ground is given by h(t) = -15t2 + 25t + 6, where t is the time in

seconds after the object is thrown. Find the average rate of the object from

1 to 4 seconds.

Answer 1

- 44 feets

Explanation:

Given the model:

h(t) = -15t2 + 25t + 6

Average height of object from 1 - 4 seconds

At :

t = 1

h(1) = -15(1)^2 + 25(1) + 6 = 16 feets

t = 2

h(2) = -15(2)^2 + 25(2) + 6 = - 4 feets

t = 3

h(3) = -15(3)^2 + 25(3) + 6 = - 54 feets

t = 4

h(4) = -15(4)^2 + 25(4) + 6 = -134 feets

Average height:

(16 + (-4) + (-54) + (-134)) / 4 feets = (- 176 / 4) feets

= - 44 feets