Question

A rocket is shot from an underground bunker. The height of the rocket after t seconds is given by y= -16t^2 + 176t -384 (measured in feet relative to the surface). Determine how long after the rocket is launched that is will emerge from the ground, and determine how much longer it will be for the rocket to hit the ground.

Answer 1

Solving the quadratic equation, it is found that:

• It emerges from the ground after 3 seconds.

• 5 seconds after that, that is, 8 seconds after being launched, it hits the ground.

What is a quadratic function?

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

The solutions are:

$x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In which:

$\Delta = b^2 - 4ac$

If a < 0, the graph, which is a parabola, is concave down, that is, it is positive only between the two roots.

In this problem, the equation is:

$y = -16t^2 + 176t - 384$

Then, the solution is:

$-16t^2 + 176t - 384 = 0$

The coefficients are $a = -16, b = 176, c = -384$, then:

$\Delta = (176)^2 - 4(-16)(-384) = 6400$

$x_1 = \frac{-176 + \sqrt{6400}}{-32} = 3$

$x_2 = \frac{-176 - \sqrt{6400}}{-32} = 8$

Considering that the parabola is concave down, we have that:

• It emerges from the ground after 3 seconds.

• 5 seconds after that, that is, 8 seconds after being launched, it hits the ground.

You can learn more about quadratic equations at https://brainly.com/question/24764843