The formula to determine the height of the rocket above the ground at any time during the rocket’s flight is given by:
`h=119t-7t2` where t =the time, in seconds, that has elapsed since the rocket was launched, and h = the height, in meters, of the rocket above the ground at time t .
The value of h was 0 when the rocket hit the ground, How many seconds after the rocket was launched did it hit the ground?

Options

A. 17 sec
B. 10 sec
C. 7 sec
D. 0 sec

Answer

To determine how many seconds after the rocket was launched it hit the ground, we need to find the value of \( t \) when the height \( h \) is equal to 0. The formula given for the height of the rocket is:

\[ h = 119t - 7t^2 \]

### Step 1: Set the height equation to zero
Since we want to find out when the rocket hits the ground, we set \( h \) to 0:

\[ 0 = 119t - 7t^2 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ 7t^2 - 119t = 0 \]

### Step 3: Factor the equation
We can factor out \( t \) from the equation:

\[ t(7t - 119) = 0 \]

### Step 4: Solve for \( t \)
Setting each factor equal to zero gives us:

1. \( t = 0 \)
2. \( 7t - 119 = 0 \)

For the second equation, we solve for \( t \):

\[ 7t = 119 \]
\[ t = \frac{119}{7} = 17 \]

### Step 5: Conclusion
The two solutions we found are \( t = 0 \) and \( t = 17 \). Since \( t = 0 \) represents the time of launch, the rocket hits the ground after:

\[ \text{The rocket hits the ground after } 17 \text{ seconds.} \]