To find out when the frisbee hits the ground, we need to determine when the height \( h(t) \) equals zero. This means we need to solve the equation:
\[ h(t) = -0.145t^2 + 0.019t + 5.5 = 0 \]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
- \( a = -0.145 \)
- \( b = 0.019 \)
- \( c = 5.5 \)
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's plug in the values for \( a \), \( b \), and \( c \):
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (0.019)^2 = 0.000361 \] \[ 4ac = 4 \cdot (-0.145) \cdot 5.5 = -3.19 \]
Now, the discriminant is:
\[ 0.000361 - (-3.19) = 0.000361 + 3.19 = 3.190361 \]
- Now we compute \( t \):
\[ t = \frac{-0.019 \pm \sqrt{3.190361}}{2 \cdot -0.145} \]
Calculating \( \sqrt{3.190361} \):
\[ \sqrt{3.190361} \approx 1.786 \]
Now, substituting back:
\[ t = \frac{-0.019 \pm 1.786}{-0.29} \]
Now we will calculate both possible values for \( t \):
- For \( t_1 = \frac{-0.019 + 1.786}{-0.29} \):
\[ t_1 = \frac{1.767}{-0.29} \approx -6.09 \quad (\text{not a valid time since it's negative}) \]
- For \( t_2 = \frac{-0.019 - 1.786}{-0.29} \):
\[ t_2 = \frac{-1.805}{-0.29} \approx 6.21 \]
Rounding to one decimal place, we get \( t \approx 6.2 \) seconds.
Thus, the frisbee will hit the ground at approximately 6.2 seconds. Therefore, the correct answer is:
6.2 seconds.