To solve for the time it takes for the basketball to reach a height of 10 feet after being thrown, we will start with the height equation provided:
\[ h = -16t^2 + 30t + 6 \]
We want to find the time \( t \) when the height \( h \) is 10 feet. Therefore, we set up the equation:
\[ 10 = -16t^2 + 30t + 6 \]
Next, we can rearrange this equation to bring all terms to one side:
\[ -16t^2 + 30t + 6 - 10 = 0 \]
This simplifies to:
\[ -16t^2 + 30t - 4 = 0 \]
To make calculations easier, we can multiply the whole equation by -1:
\[ 16t^2 - 30t + 4 = 0 \]
Now, we can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 16 \), \( b = -30 \), and \( c = 4 \). We start by calculating the discriminant:
\[ b^2 - 4ac = (-30)^2 - 4 \cdot 16 \cdot 4 \] \[ = 900 - 256 \] \[ = 644 \]
Now, we plug in the values into the quadratic formula:
\[ t = \frac{-(-30) \pm \sqrt{644}}{2 \cdot 16} \] \[ = \frac{30 \pm \sqrt{644}}{32} \]
Next, we find \( \sqrt{644} \):
\[ \sqrt{644} = 2\sqrt{161} \]
Then we can calculate \( t \):
\[ t = \frac{30 \pm 2\sqrt{161}}{32} \]
To simplify further:
\[ t = \frac{30}{32} \pm \frac{2\sqrt{161}}{32} \] \[ = \frac{15}{16} \pm \frac{\sqrt{161}}{16} \]
Thus, we have two potential solutions:
\[ t_1 = \frac{15 + \sqrt{161}}{16} \] \[ t_2 = \frac{15 - \sqrt{161}}{16} \]
Now let's estimate \( \sqrt{161} \):
\[ \sqrt{161} \approx 12.6886 \quad (\text{since } 12.6886^2 \approx 161) \]
Calculating \( t_1 \) and \( t_2 \):
- For \( t_1 \):
\[ t_1 \approx \frac{15 + 12.6886}{16} \approx \frac{27.6886}{16} \approx 1.733 \text{ seconds} \]
- For \( t_2 \):
\[ t_2 \approx \frac{15 - 12.6886}{16} \approx \frac{2.3114}{16} \approx 0.144 \text{ seconds} \]
Since the ball is thrown upward and then comes back down, we only consider the value of \( t_1 \). Therefore, the time it takes for the basketball to go into the hoop at a height of 10 feet is approximately:
\[ \boxed{1.73 \text{ seconds}} \]
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