Question

fireworks are shot by remote - control into the air from a pit that is 9 ft below earths surface.
(a) find an equation that models the height of a roman candle t seconds after it is shot upward with an initial velocity of 60 ft/sec. graph the equation.
(b) what is the maximum height above ground - level that the roman candle will reach? how many seconds will it take to reach that height?
graph the equation. choose the correct graph below.
0, 5 by -9,108 0, 5 by -9,108 0, 5 by -9,108 0, 5 by -9,108
(b) what is the maximum height above ground - level that the roman candle will reach? how many seconds will it take to reach that height?
the roman candle will reach the maximum height of in
(simplify your answers.)

Explanation:

Step1: Recall the height - velocity formula

The general formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $v_{0}$ is the initial velocity and $h_{0}$ is the initial height. The pit is 9 ft below the Earth's surface, so $h_{0}=- 9$ ft and $v_{0}=80$ ft/sec. Thus, the equation is $h(t)=-16t^{2}+80t - 9$.

Step2: Find the time to reach the maximum height

The function $h(t)=-16t^{2}+80t - 9$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 80$, and $c=-9$. The time $t$ at which the object reaches its maximum height is given by the formula $t=-\frac{b}{2a}$. Substituting the values of $a$ and $b$ into the formula, we have $t =-\frac{80}{2\times(-16)}=\frac{80}{32}=\frac{5}{2}=2.5$ seconds.

Step3: Find the maximum height

Substitute $t = 2.5$ into the height function $h(t)=-16t^{2}+80t - 9$.
$h(2.5)=-16\times(2.5)^{2}+80\times2.5-9$.
First, calculate $-16\times(2.5)^{2}=-16\times6.25=-100$.
Then, $80\times2.5 = 200$.
So, $h(2.5)=-100 + 200-9=91$ ft.

Answer:

The Roman candle will reach the maximum height of 91 ft in 2.5 seconds.