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A toy rocket is launched straight up from the ground with an initial velocity of [tex]$80 \, \text{ft/s}$[/tex] and returns to the ground after 5 seconds. The height of the rocket [tex]t[/tex] seconds after launch is modeled by the function [tex]f(t) = -16t^2 + 80t[/tex].

What is the maximum height of the rocket, in feet?

A. 100
B. 80
C. 16
D. 50

Answer :

To find the maximum height of the toy rocket, we need to analyze the function modeling its height:

The function given is [tex]\( f(t) = -16t^2 + 80t \)[/tex], where [tex]\( t \)[/tex] is the time in seconds after the rocket is launched.

This function is quadratic, and the graph of a quadratic function is a parabola. Because the coefficient of [tex]\( t^2 \)[/tex] is negative (-16), the parabola opens downwards, meaning it has a maximum point (the vertex).

To find the time at which the maximum height occurs, we use the formula for the vertex of a parabola, [tex]\( t = -\frac{b}{2a} \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the coefficients from the quadratic function:

- Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 80 \)[/tex].

Plug these values into the formula:

[tex]\[
t = -\frac{80}{2 \times -16} = \frac{80}{32} = 2.5
\][/tex]

So, the rocket reaches its maximum height at [tex]\( t = 2.5 \)[/tex] seconds.

Now, to find the maximum height, substitute [tex]\( t = 2.5 \)[/tex] back into the height function:

[tex]\[
f(2.5) = -16(2.5)^2 + 80(2.5)
\][/tex]

Calculate each part:

1. [tex]\( (2.5)^2 = 6.25 \)[/tex]
2. [tex]\( -16 \times 6.25 = -100 \)[/tex]
3. [tex]\( 80 \times 2.5 = 200 \)[/tex]

Adding these together gives:

[tex]\[
-100 + 200 = 100
\][/tex]

Therefore, the maximum height of the rocket is 100 feet.

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