To find the values of \( t \) for which the rocket's height \( H \) is 34 feet, we start with the given equation for height:
\[ H = 100t - 16t^2 \]
We need to set \( H \) equal to 34 feet:
\[ 34 = 100t - 16t^2 \]
Rearranging this equation gives us:
\[ 16t^2 - 100t + 34 = 0 \]
Next, we can apply the quadratic formula, which is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 16 \), \( b = -100 \), and \( c = 34 \). First, we can calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-100)^2 - 4 \cdot 16 \cdot 34 \] \[ b^2 - 4ac = 10000 - 2176 \] \[ b^2 - 4ac = 7824 \]
We can use this in the quadratic formula to find \( t \):
\[ t = \frac{-(-100) \pm \sqrt{7824}}{2 \cdot 16} \] \[ t = \frac{100 \pm \sqrt{7824}}{32} \]
Calculating \( \sqrt{7824} \):
\[ \sqrt{7824} \approx 88.49 \]
Substituting back, we get:
\[ t = \frac{100 \pm 88.49}{32} \]
This yields two values for \( t \):
\[ t_1 = \frac{100 + 88.49}{32} = \frac{188.49}{32} \approx 5.89 \]
\[ t_2 = \frac{100 - 88.49}{32} = \frac{11.51}{32} \approx 0.36 \]
Therefore, the two values of \( t \) for which the height of the rocket is 34 feet are:
\[ t \approx 0.36 \text{ seconds or } t \approx 5.89 \text{ seconds} \]
So, the final rounded answers are:
t ≈ 0.36 or t ≈ 5.89.