Question

1. 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)=-16t² + 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.

Explanation:

Step1: Find the time of maximum height

For a quadratic function $y = ax^{2}+bx + c$, the $x$-coordinate (in this case $t$ - coordinate) of the vertex is $t=-\frac{b}{2a}$. Here, $a=-16$, $b = 120$, so $t=-\frac{120}{2\times(-16)}=\frac{120}{32}=\frac{15}{4}=3.75$ seconds.

Step2: Find the initial height

When $t = 0$, $h(0)=-16\times0^{2}+120\times0 + 6=6$ feet.

Step3: Find the height at maximum - time

Substitute $t = 3.75$ into $h(t)=-16t^{2}+120t + 6$.
$h(3.75)=-16\times(3.75)^{2}+120\times3.75 + 6=-16\times14.0625+450 + 6=-225+450 + 6=231$ feet.

Step4: Calculate the average rate of change

The average rate of change formula is $\frac{h(t_2)-h(t_1)}{t_2 - t_1}$. Here, $t_1 = 0$, $t_2=3.75$, $h(t_1)=6$, $h(t_2)=231$.
$\frac{h(3.75)-h(0)}{3.75 - 0}=\frac{231 - 6}{3.75}=\frac{225}{3.75}=60$ feet per second.

Answer:

60 feet per second