11.

Question

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

A rocket is launched from atop a 55-foot cliff with an initial velocity of 138 ft/s.
a. Substitute the values into the vertical motion formula h = −16t2 + vt + c. Let h = 0.
b. Use the quadratic formula find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.

0 = −16t2 + 55t + 138; 0.4 s

0 = −16t2 + 55t + 138; 9 s

0 = −16t2 + 138t + 55; 0.4 s

0 = −16t2 + 138t + 55; 9 s

Answer 1

To solve this problem, we'll use the vertical motion formula given by:

\[ h = -16t^2 + vt + c \]

In this formula, \( h \) represents the height, \( t \) is the time in seconds, \( v \) is the initial velocity, and \( c \) is the initial height from which the object is launched. We need to find out when \( h = 0 \), which means the rocket hits the ground.

We are given:
- Initial velocity \( v = 138 \, \text{ft/s} \)
- Initial height \( c = 55 \, \text{ft} \)

The equation for the height becomes:
\[ 0 = -16t^2 + 138t + 55 \]

Now, let's solve \( -16t^2 + 138t + 55 = 0 \) using the quadratic formula. The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the equation:
\[ 0 = -16t^2 + 138t + 55 \]
- \( a = -16 \)
- \( b = 138 \)
- \( c = 55 \)

First, calculate the discriminant:
\[ \Delta = b^2 - 4ac = 138^2 - 4(-16)(55) \]
\[ \Delta = 19044 + 3520 \]
\[ \Delta = 22564 \]

Now, apply the quadratic formula:
\[ t = \frac{-138 \pm \sqrt{22564}}{2(-16)} \]
\[ t = \frac{-138 \pm 150.21}{-32} \]

We get two solutions:
\[ t = \frac{-138 + 150.21}{-32} = \frac{12.21}{-32} = -0.4 \]
\[ t = \frac{-138 - 150.21}{-32} = \frac{-288.21}{-32} = 9 \]

Since time cannot be negative, we discard \( t = -0.4 \) and take \( t = 9 \) seconds.

Thus, the rocket will take \( 9 \) seconds to hit the ground after it is launched.

The correct answer is:
\[ \boxed{0 = -16t^2 + 138t + 55; 9 \, s} \]