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 $10.17\ seconds$.

Answer 2

Time can not be negative from any perspective. Hence, the correct answer is 10.16 seconds.

The equation of the path of a rocket is given that is $y=-16x^2+153x+98$

We need to determine the time that the rocket will hit the ground.

Now, if the rocket hits the ground after the launching then the overall displacement at the time of hitting the ground will be zero.

Therefore, the value of y is 0.

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

Now, the formula for finding the roots of the quadratic equation $ax^2+bx+c=0$ is given as:

$\begin{aligned}x_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\end{aligned}$

Comparing the expression with the standard form of the equation.

$a=-16\;b=153\;\rm{and}\;c=98$

By applying the formula there are two values are come that is,

$x_1=-0.602555\\x_2=10.1651$

Time can not be negative from any perspective.

Hence, the correct answer is 10.16 seconds.

To know more about it, please refer to the link:

https://brainly.com/question/18389900