To find out when the ball hits the ground, we need to determine when the height \( h(t) = 0 \). The height of the ball is given by the equation:
\[ h(t) = 44 - 6t - 5t^2 \]
Setting this equal to zero to find when the ball hits the ground:
\[ 44 - 6t - 5t^2 = 0 \]
Rearranging the equation gives us:
\[ -5t^2 - 6t + 44 = 0 \]
Multiplying the entire equation by -1 to make the leading coefficient positive:
\[ 5t^2 + 6t - 44 = 0 \]
Now we will use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \( a = 5 \), \( b = 6 \), and \( c = -44 \). We first calculate the discriminant:
\[ b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-44) = 36 + 880 = 916 \]
Now using the quadratic formula:
\[ t = \frac{-6 \pm \sqrt{916}}{2 \cdot 5} \]
Calculating \( \sqrt{916} \):
\[ \sqrt{916} \approx 30.233 \]
Now substituting back into the formula:
\[ t = \frac{-6 \pm 30.233}{10} \]
Calculating the two possible values for \( t \):
- Using the plus sign:
\[ t = \frac{-6 + 30.233}{10} = \frac{24.233}{10} \approx 2.4233 \]
- Using the minus sign:
\[ t = \frac{-6 - 30.233}{10} = \frac{-36.233}{10} \approx -3.6233 \quad (\text{not physically meaningful, as time cannot be negative}) \]
Thus, the time when the ball hits the ground is:
\[ t \approx 2.42 \text{ seconds (rounded to the nearest hundredth)} \]
The answer is:
2.42 seconds.