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.


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

Answer 1

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 \( x \):

1. First value:

\[
x_1 = \frac{-118 + 134.86}{-32} = \frac{16.86}{-32} \approx -0.5275 \text{ (not a valid solution, as time cannot be negative)}
\]

2. 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:

\[
\boxed{7.91 \text{ seconds}}
\]