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Genrish500
8 days ago
11

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.
Mathematics
1 answer:
lawyer [4K]8 days ago
6 0

Answer:

60

Step-by-step explanation:

The function provided is:

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

The average rate of change of h(t) as time goes from t=a to t=b is expressed as:

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

This function can be reformulated as: h(t)=-16(t-3.75)^2+231

The rocket's peak height is 231, which occurs at t=3.75 seconds.

\implies h(3.75)=231

The initial launch happens at: t=0

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

The average rate of change from launch to max height is

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

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