The given equation is:

$y = -16x^2 + 180x + 93$

Here, $y$ represents the height of the rocket in feet, and $x$ represents the time in seconds. To find the time when the rocket hits the ground, we need to find when the height $y$ becomes 0 (i.e., when the rocket reaches the ground):

$0 = -16x^2 + 180x + 93$

This is a quadratic equation, and we can solve for $x$ (time) using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $-16x^2 + 180x + 93 = 0$, the coefficients are:

• $a = -16$
• $b = 180$
• $c = 93$

Substitute these values into the quadratic formula to solve for $x$.

I'll calculate it now.The two solutions for the time $x$ are approximately:

• $x_1 = -0.49$ (which is not physically meaningful, as time cannot be negative),
• $x_2 = 11.74$ seconds.

Therefore, the rocket will hit the ground approximately 11.74 seconds after launch.

Would you like further details on the steps involved or any clarifications?

Here are some related questions:

1. How does the discriminant of a quadratic equation determine the nature of the solutions?
2. What is the physical meaning of the coefficients in the rocket's height equation?
3. How can we derive the maximum height of the rocket from the given equation?
4. How does changing the initial velocity affect the time to hit the ground?
5. Can we model other types of motion (e.g., free fall) using a similar equation?

Tip: When solving quadratic equations, always check if a negative solution makes sense in the context of the problem.