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.

Answer 1

Here is the full 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+153x+98

Answer:

10.2 seconds

Step-by-step explanation:

Given equation :

$y=-16x^2+153x+98$ shown the expression of a quadratic equation

Let y =0

∴

0 = $-16x^2+153x+98$

where;

$a=-16\\b=153\\c=98$

Using the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

replacing it with our values; we have:

$x=\frac{-153\±\sqrt{153^2-4(-16)(98)}}{2(-16)}$

$x=\frac{-153\ + \sqrt{153^2-4(-16)(98)}}{2(-16)}$    OR     $x=\frac{-153\ - \sqrt{153^2-4(-16)(98)}}{2(-16)}$

$x_1=10.17$   OR     $x_2=-0.60$

Hence, we go by the positive value since,  is the time that the rocket will hit the ground.

x= 10.17

x = ≅ 10.2 seconds

Therefore, the rocket will hit the ground,  to the nearest 100th of second.  = 10.2 seconds