A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket , to the nearest 100th of a foot. y=-16x^2+230x+112

Respuesta :

Answer:

The maximum height reached by the rocket is of 938.56 feet.

Step-by-step explanation:

The height y, after x seconds, is given by a equation in the following format:

[tex]y(x) = ax^{2} + bx + c[/tex]

If a is negative, the maximum height is:

[tex]y(x_{v})[/tex]

In which

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

In this question:

[tex]y(x) = -16x^{2} + 230x + 112[/tex]

So

[tex]a = -16, b = 230, c = 112[/tex]

Then

[tex]x_{v} = -\frac{230}{2*(-16)} = 7.1875[/tex]

[tex]y(7.1835) = -16*(7.1835)^{2} + 230*7.1835 + 112 = 938.56[/tex]

The maximum height reached by the rocket is of 938.56 feet.