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A toy rocket is shot vertically into the air from a launching pad 4 feet above the ground with an initial velocity of 40 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 + 40t + 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 the quadratic function

[tex]$$
h(t) = -16t^2 + 40t + 4.
$$[/tex]

For a quadratic function of the form

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

the maximum (or minimum) value occurs at

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

Here, [tex]$a = -16$[/tex] and [tex]$b = 40$[/tex]. Thus, the time at which the rocket reaches its maximum height is

[tex]$$
t = -\frac{40}{2(-16)} = 1.25 \text{ seconds}.
$$[/tex]

To find the maximum height, we substitute [tex]$t = 1.25$[/tex] back into the height function:

[tex]$$
h(1.25) = -16(1.25)^2 + 40(1.25) + 4.
$$[/tex]

Calculating step by step:

1. Compute [tex]$(1.25)^2$[/tex]:
[tex]$$
(1.25)^2 = 1.5625.
$$[/tex]
2. Multiply by [tex]$-16$[/tex]:
[tex]$$
-16 \times 1.5625 = -25.
$$[/tex]
3. Multiply [tex]$40$[/tex] by [tex]$1.25$[/tex]:
[tex]$$
40 \times 1.25 = 50.
$$[/tex]
4. Sum the results along with [tex]$4$[/tex]:
[tex]$$
-25 + 50 + 4 = 29.
$$[/tex]

Thus, the maximum height reached by the rocket is [tex]$29$[/tex] feet.

The final answers are:
- The rocket reaches its maximum height [tex]$1.25$[/tex] second(s) after launch.
- The maximum height is [tex]$29$[/tex] feet.

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Rewritten by : Barada