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A toy rocket is launched vertically into the air from a pad 4 feet above the ground with an initial velocity of 80 feet per second. The height [tex] h [/tex], in feet, of the rocket above the ground at [tex] t [/tex] seconds after launch is given by the function:

[tex] h(t) = -16t^2 + 80t + 4 [/tex]

1. How long will it take the rocket to reach its maximum height?
2. What is the maximum height?

The rocket reaches its maximum height at [tex] \square [/tex] second(s) after launch. (Simplify your answer.)

Answer :

The height of the rocket is given by

[tex]$$
h(t) = -16t^2 + 80t + 4.
$$[/tex]

Since this is a quadratic function and the coefficient of [tex]$t^2$[/tex] is negative, the graph is a downward-opening parabola. The maximum height occurs at the vertex of this parabola.

For a quadratic function of the form

[tex]$$
at^2 + bt + c,
$$[/tex]

the [tex]$t$[/tex]-coordinate of the vertex is found using

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

Here, [tex]$a = -16$[/tex] and [tex]$b = 80$[/tex], so the time at which the rocket reaches its maximum height is

[tex]$$
t = -\frac{80}{2(-16)} = 2.5 \text{ seconds}.
$$[/tex]

To find the maximum height, substitute [tex]$t = 2.5$[/tex] into the height function:

[tex]$$
h(2.5) = -16(2.5)^2 + 80(2.5) + 4.
$$[/tex]

First, calculate [tex]$(2.5)^2$[/tex]:

[tex]$$
(2.5)^2 = 6.25.
$$[/tex]

Then, compute each term:

[tex]$$
-16 \times 6.25 = -100, \quad 80 \times 2.5 = 200.
$$[/tex]

Now, sum the terms:

[tex]$$
h(2.5) = -100 + 200 + 4 = 104 \text{ feet}.
$$[/tex]

Thus, the rocket reaches its maximum height of [tex]$104$[/tex] feet at [tex]$\boxed{2.5}$[/tex] seconds after launch.

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