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

Question

3 months ago

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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 [12.5K] · Answer 1 · 2026-02-24T22:18:55+03:00

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

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