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.
Accepted Solution
A:
Answer:60Step-by-step explanation:The given function is:[tex]h(t)=-16t^2+120t+6[/tex]The average rate of change of h(t) from t=a to t=b is given by:[tex]\frac{h(b)-h(a)}{b-a}[/tex]We can rewrite this function as: [tex]h(t)=-16(t-3.75)^2+231[/tex]The maximum height of the rocket is 231 and it occurs at t=3.75[tex]\implies h(3.75)=231[/tex]The initial launch occurs at: t=0and [tex]h(0)=-16(0)^2+120(0)+6=6[/tex]The average rate of change from the initial launch to the maximum height is [tex]\frac{h(3.75)-h(0)}{3.75-0}=\frac{231-6}{3.75-0} =60[/tex]