To find how long after the ball is thrown it hits the ground, we need to solve the equation for \( h = 0 \):
\[ 217 - 17t - 16t^2 = 0 \]
Rearranging this gives us:
\[ -16t^2 - 17t + 217 = 0 \]
Multiplying the entire equation by -1 to make it standard form:
\[ 16t^2 + 17t - 217 = 0 \]
Now, we can use the quadratic formula to solve for \( t \). The quadratic formula is given by:
\[ t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
In our case, \( a = 16 \), \( b = 17 \), and \( c = -217 \). First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 17^2 - 4 \cdot 16 \cdot (-217) \]
Calculating \( b^2 \):
\[ 17^2 = 289 \]
Calculating \( 4ac \):
\[ 4 \cdot 16 \cdot (-217) = -13952 \]
Now putting it together:
\[ b^2 - 4ac = 289 + 13952 = 14241 \]
Now we can find \( \sqrt{14241} \):
\[ \sqrt{14241} \approx 119.34 \quad (\text{using a calculator}) \]
Putting it back into the quadratic formula:
\[ t = \frac{{-17 \pm 119.34}}{{2 \cdot 16}} \]
Calculating the two potential solutions for \( t \):
- First solution:
\[ t = \frac{{-17 + 119.34}}{{32}} = \frac{{102.34}}{{32}} \approx 3.20 \]
- Second solution:
\[ t = \frac{{-17 - 119.34}}{{32}} = \frac{{-136.34}}{{32}} \approx -4.26 \]
Since time cannot be negative, the solution that makes sense is:
\[ t \approx 3.20 \text{ seconds} \]
Thus, the ball hits the ground approximately 3.20 seconds after it was thrown.