Question

during the late ming dynasty in china (16th century), a legendary official named wan hu achieved the earliest known rocket flight four hundred years before the wright brothers made history at kitty hawk.
wan hus goal was to launch himself to outer space using chinas advanced rocket and fireworks technology. his launch vehicle was a wooden chair built with forty - seven rockets attached. on the day of lift - off, wan, splendidly attired, climbed into his rocket chair and forty seven servants lit the fuses and then hastily ran for cover. there was a huge explosion. when the smoke cleared, wan and the chair were gone, and was said never to have been seen again.
although this early attempt at space flight was not repeated, wan hu left behind several aeronautical manuscripts detailing his calculations. what follows below is the equation of motion for his trajectory recovered from those manuscripts (converted into our modern units), where his vertical position is given as h (feet) as a function of time t (seconds):
h(t)=-16t² + 886.4t
wan hu estimated that outer space could be reached at an altitude (in our modern units) of
use your knowledge of the quadratic formula to calculate the interval of time during which w expected to be in outer space. round your time values to two decimals and enter in the form
time interval: seconds
rewrite wan hus equation of motion in vertex form, and estimate his maximum altitude hmax. round all numbers to two decimal places:
h(t)=
hmax= feet

Explanation:

Step1: Find time - interval for outer - space

Set $h(t)$ equal to the outer - space altitude (not given in the problem, assume we want to find when $h(t)\geq0$). The quadratic function is $h(t)=-16t^{2}+886.4t$. Set $h(t) = 0$ and solve for $t$ using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a=-16$, $b = 886.4$, and $c = 0$.
$t=\frac{-886.4\pm\sqrt{886.4^{2}-4\times(-16)\times0}}{2\times(-16)}=\frac{-886.4\pm886.4}{-32}$.
The two solutions are $t_1 = 0$ and $t_2=\frac{-886.4 - 886.4}{-32}=\frac{-1772.8}{-32}=55.40$. The time interval when $h(t)\geq0$ is $[0,55.40]$.

Step2: Rewrite in vertex - form

The vertex - form of a quadratic function $y = ax^{2}+bx + c$ is $y=a(x - h)^{2}+k$, where $h=-\frac{b}{2a}$ and $k = y(h)$.
For $h(t)=-16t^{2}+886.4t$, $a=-16$, $b = 886.4$. First, find $h$: $h=-\frac{886.4}{2\times(-16)}=\frac{886.4}{32}=27.70$.
Then find $k$: $k=h(27.70)=-16\times(27.70)^{2}+886.4\times27.70=-16\times767.29+24553.28=-12276.64 + 24553.28=12276.64$.
The vertex - form is $h(t)=-16(t - 27.70)^{2}+12276.64$.
The maximum altitude occurs at the vertex, so $h_{max}=12276.64$ feet.

Answer:

time interval: $[0,55.40]$
$h(t)=-16(t - 27.70)^{2}+12276.64$
$h_{max}=12276.64$ feet