The equation given for the height of the rocket is:

$y = -16x^2 + 118x + 69$

To find the time when the rocket hits the ground, we set $y = 0$, since the height of the rocket is 0 when it reaches the ground. Therefore, we solve the equation:

$0 = -16x^2 + 118x + 69$

This is a quadratic equation. We can solve it using the quadratic formula:

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

Where:

• $a = -16$
• $b = 118$
• $c = 69$

Substitute the values into the quadratic formula:

$x = \frac{-118 \pm \sqrt{118^2 - 4(-16)(69)}}{2(-16)}$

Let's calculate the discriminant first:

$118^2 = 13924$ $4(-16)(69) = -4416$ $\text{Discriminant} = 13924 + 4416 = 18340$

Now, substitute the discriminant back into the quadratic formula:

$x = \frac{-118 \pm \sqrt{18340}}{-32}$

$x = \frac{-118 \pm 135.33}{-32}$

Now solve for both values of $x$:

1. $x = \frac{-118 + 135.33}{-32} = \frac{17.33}{-32} = -0.542$
2. $x = \frac{-118 - 135.33}{-32} = \frac{-253.33}{-32} = 7.92$

Since time cannot be negative, we take the positive root. So, the rocket will hit the ground after approximately 7.92 seconds.

Would you like more details on the quadratic formula or any other questions?

Related questions:

1. How does the discriminant affect the nature of the roots in a quadratic equation?
2. What does the term $-16$ in the height equation represent physically?
3. How can we graph this quadratic equation to visualize the rocket's trajectory?
4. If the rocket had a different initial velocity, how would the equation change?
5. How can we use the vertex form of the quadratic to determine the rocket's maximum height?

Tip: Always remember to check the discriminant when solving quadratic equations; it tells you if the roots are real, complex, or repeated!