To find the values of \( t \) for which the rocket's height \( h(t) = 140t - 16t^2 \) is equal to 68 feet, we can set up the equation:
\[ 140t - 16t^2 = 68 \]
Rearranging this gives us:
\[ -16t^2 + 140t - 68 = 0 \]
To make it easier to solve, we can multiply the entire equation by -1:
\[ 16t^2 - 140t + 68 = 0 \]
Next, we can use the quadratic formula to solve for \( t \) where \( a = 16 \), \( b = -140 \), and \( c = 68 \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant \( b^2 - 4ac \):
\[ b^2 = (-140)^2 = 19600 \] \[ 4ac = 4 \times 16 \times 68 = 4352 \] \[ b^2 - 4ac = 19600 - 4352 = 15248 \]
Now we can find the square root of the discriminant:
\[ \sqrt{15248} \approx 123.46 \]
Now substituting back into the quadratic formula:
\[ t = \frac{140 \pm 123.46}{32} \]
Calculating the two possible values for \( t \):
- For the positive root:
\[ t = \frac{140 + 123.46}{32} \approx \frac{263.46}{32} \approx 8.24 \]
- For the negative root:
\[ t = \frac{140 - 123.46}{32} \approx \frac{16.54}{32} \approx 0.52 \]
Thus, the two times \( t \) when the rocket is at a height of 68 feet are approximately:
\[ t \approx 0.52 \quad \text{and} \quad t \approx 8.24 \]
Therefore, rounding to the nearest hundredth, the final answers are:
\[ t \approx 0.52 \quad \text{or} \quad t \approx 8.24 \]