Question

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

Answer 1

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

Explanation:

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

`y(x) = ax^(2) + bx + c`

If a is negative, the maximum height is:

`y(x_(v))`

In which

`x_(v) = -(b)/(2a)`

In this question:

`y(x) = -16x^(2) + 230x + 112`

So

`a = -16, b = 230, c = 112`

Then

`x_(v) = -(230)/(2*(-16)) = 7.1875`

`y(7.1835) = -16*(7.1835)^(2) + 230*7.1835 + 112 = 938.56`

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