Question

A rocket was launched into the air from a podium 6 feet off the ground. The rocket path is represented by the equation h(t)=-16t^2+120t+6, where h(t) represents the height, in feet, and t is the time, in seconds. Find the average rate of change from the initial launch to the maximum height.

Answer 1

Answer:

60

Step-by-step explanation:

The given function is:

$h(t)=-16t^2+120t+6$

The average rate of change of h(t) from t=a to t=b is given by:

$\frac{h(b)-h(a)}{b-a}$

We can rewrite this function as: $h(t)=-16(t-3.75)^2+231$

The maximum height of the rocket is 231 and it occurs at t=3.75

$\implies h(3.75)=231$

The initial launch occurs at: t=0

and $h(0)=-16(0)^2+120(0)+6=6$

The average rate of change from the initial launch to the maximum height is

$\frac{h(3.75)-h(0)}{3.75-0}=\frac{231-6}{3.75-0} =60$