Mikalee, a fighter pilot, ejects from her burning plane. Once she is safely on the ground, she launches a flare to attract the attention of a rescue plane. The initial speed of the flare is [tex]$1600 \text{ ft/sec}$[/tex]. The height of the flare, [tex]h(t)[/tex], at any time, [tex]t[/tex], is modeled by the function [tex]h(t) = -16t^2 + 1600t[/tex].

What is the maximum height of the flare?



Answer :

To determine the maximum height of the flare launched by Mikalee, we need to analyze the function that describes the height [tex]\( h(t) \)[/tex] of the flare over time [tex]\( t \)[/tex]. The height of the flare is given by the quadratic function:

[tex]\[ h(t) = -16t^2 + 1600t \][/tex]

This quadratic function is a parabola that opens downwards, which means it has a maximum point, also known as the vertex of the parabola. The vertex formula for a parabola given by [tex]\( ax^2 + bx + c \)[/tex] is found using [tex]\( t = -\frac{b}{2a} \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the coefficients from the given quadratic equation.

Let's identify the coefficients:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 1600 \)[/tex]

Now we use the vertex formula:

[tex]\[ t = -\frac{b}{2a} \][/tex]

Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

[tex]\[ t = -\frac{1600}{2(-16)} \][/tex]
[tex]\[ t = -\frac{1600}{-32} \][/tex]
[tex]\[ t = 50 \][/tex]

So, the flare reaches its maximum height at [tex]\( t = 50 \)[/tex] seconds.

Next, to find the maximum height [tex]\( h_{\text{max}} \)[/tex], we substitute [tex]\( t = 50 \)[/tex] back into the original height function [tex]\( h(t) \)[/tex]:

[tex]\[ h(50) = -16(50)^2 + 1600(50) \][/tex]

First, calculate [tex]\( 50^2 \)[/tex]:

[tex]\[ 50^2 = 2500 \][/tex]

Now substitute [tex]\( 2500 \)[/tex] and simplify:

[tex]\[ h(50) = -16(2500) + 1600(50) \][/tex]
[tex]\[ h(50) = -40000 + 80000 \][/tex]
[tex]\[ h(50) = 40000 \][/tex]

Thus, the maximum height of the flare is [tex]\( 40000 \)[/tex] feet.