To determine where the water hits the ground, we need to find the point at which \(y = 0\) (the ground level) in the given equation:
\[ y = -34(x - 2)^2 + 3 \]
Setting \(y = 0\):
\[ 0 = -34(x - 2)^2 + 3 \]
Now we will solve for \(x\):
- Subtract 3 from both sides:
\[ -3 = -34(x - 2)^2 \]
- Divide by -34:
\[ \frac{3}{34} = (x - 2)^2 \]
- Take the square root of both sides, remembering to consider both the positive and negative roots:
\[ x - 2 = \pm \sqrt{\frac{3}{34}} \]
- Solve for \(x\):
\[ x = 2 \pm \sqrt{\frac{3}{34}} \]
Next, we can calculate the approximate values:
\[ \sqrt{\frac{3}{34}} \approx \sqrt{0.08824} \approx 0.297 \]
Thus, we can find the two potential \(x\) values where the water hits the ground:
\[ x \approx 2 + 0.297 \quad \text{and} \quad x \approx 2 - 0.297 \]
Calculating these:
- \(x \approx 2 + 0.297 \approx 2.297\)
- \(x \approx 2 - 0.297 \approx 1.703\)
Therefore, the water will hit the ground at approximately:
\[ x \approx 1.703 \quad \text{and} \quad x \approx 2.297 \]
So, the points where the water hits the ground are approximately \( (1.703, 0) \) and \( (2.297, 0) \).