To find the time when the rocket will hit the ground, we need to determine when the height \( y \) is equal to 0. This means we need to solve the equation:
\[ -16x^2 + 118x + 69 = 0 \]
We can use the quadratic formula to solve for \( x \), given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = -16 \), \( b = 118 \), and \( c = 69 \). First, we need to calculate the discriminant:
\[ b^2 - 4ac = 118^2 - 4 \cdot (-16) \cdot 69 \]
Calculating \( b^2 \):
\[ 118^2 = 13924 \]
Calculating \( 4ac \):
\[ 4 \cdot (-16) \cdot 69 = -4256 \]
Now we can compute the discriminant:
\[ b^2 - 4ac = 13924 - (-4256) = 13924 + 4256 = 18180 \]
Next, we apply the quadratic formula:
\[ x = \frac{-118 \pm \sqrt{18180}}{2 \cdot -16} \]
Calculating \( \sqrt{18180} \):
\[ \sqrt{18180} \approx 134.86 \]
Now substituting this back into the formula:
\[ x = \frac{-118 \pm 134.86}{-32} \]
Calculating the two possible values for \( x \):
- First value:
\[ x_1 = \frac{-118 + 134.86}{-32} = \frac{16.86}{-32} \approx -0.5275 \text{ (not a valid solution, as time cannot be negative)} \]
- Second value:
\[ x_2 = \frac{-118 - 134.86}{-32} = \frac{-252.86}{-32} \approx 7.91 \]
Thus, the time that the rocket will hit the ground, to the nearest hundredth of a second, is:
\[ \boxed{7.91 \text{ seconds}} \]