Question

a flowerpot topples off a window ledge and falls to the street below. the height, in feet, of the flowerpot above the ground is given by the equation $h = -16t^2 + 16$, where $t$ is the number of seconds the flowerpot has been falling.

what is the average rate of change of the height of the flowerpot on the interval from $t = 0.75$ to $t = 1$?

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function \( h(t) \) on the interval \([a, b]\) is given by \(\frac{h(b)-h(a)}{b - a}\). Here, \( a = 0.75 \), \( b = 1 \), and \( h(t)=- 16t^{2}+16 \).

Step2: Calculate \( h(1) \)

Substitute \( t = 1 \) into \( h(t) \):
\( h(1)=-16(1)^{2}+16=-16 + 16=0 \)

Step3: Calculate \( h(0.75) \)

Substitute \( t = 0.75 \) into \( h(t) \):
\( h(0.75)=-16(0.75)^{2}+16=-16\times0.5625 + 16=-9 + 16 = 7 \)

Step4: Apply average rate of change formula

Using the formula \(\frac{h(1)-h(0.75)}{1 - 0.75}\), substitute the values:
\(\frac{0 - 7}{1 - 0.75}=\frac{-7}{0.25}=-28\)

Answer:

-28