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 :
[tex]y=-16x^2+153x+98[/tex] shown the expression of a quadratic equation
Let y =0
∴
0 = [tex]-16x^2+153x+98[/tex]
where;
[tex]a=-16\\b=153\\c=98[/tex]
Using the quadratic formula:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
replacing it with our values; we have:
[tex]x=\frac{-153\±\sqrt{153^2-4(-16)(98)}}{2(-16)}[/tex]
[tex]x=\frac{-153\ + \sqrt{153^2-4(-16)(98)}}{2(-16)}[/tex] OR [tex]x=\frac{-153\ - \sqrt{153^2-4(-16)(98)}}{2(-16)}[/tex]
[tex]x_1=10.17[/tex] OR [tex]x_2=-0.60[/tex]
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
