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 1

Answer: $10.17\ seconds$

Step-by-step explanation:

Given the following Quadratic equation:

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

The steps to solve it are:

1. Substitute $y=0$ into the equation:

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

2. Use the Quadratic formula $x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$.

In this case:

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

Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):

$x=\frac{-153\±\sqrt{153^2-4(-16)(98)}}{2(-16)}\\\\x_1=10.17\\x_2=-0.60$

3. The positive value is the time that the rocket will hit the ground.

Therefore, the rocket will hit the ground after $10.17\ seconds$.