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 time that the rocket will hit the ground, to the nearest 100th of second.
equation: y=−16x^2+177x+98

Answer 1

Answer:

The rocket hits the ground at a time of 11.59 seconds.

Step-by-step explanation:

The height of the rocket, after x seconds, is given by the following equation:

$y = -16x^2 + 177x + 98$

It hits the ground when $y = 0$, so we have to find x for which $y = 0$, which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$y = -16x^2 + 177x + 98$

$-16x^2 + 177x + 98 = 0$

So

$a = -16, b = 177, c = 98$

$\bigtriangleup = 177^{2} - 4(-16)(98) = 37601$

$x_{1} = \frac{-177 + \sqrt{37601}}{2*(-16)} = -0.53$

$x_{2} = \frac{-177 - \sqrt{37601}}{2*(-16)} = 11.59$

Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.