To determine when the rock hits the water, we need to find the time \( t \) when the height \( h \) equals 0 (when the rock reaches the water level).
The equation given is:
\[ h = -4.9t^2 + 10 \]
To find when it hits the water, set \( h = 0 \):
\[ 0 = -4.9t^2 + 10 \]
Now we can solve for \( t \):
- Rearrange the equation:
\[ 4.9t^2 = 10 \]
- Divide both sides by 4.9:
\[ t^2 = \frac{10}{4.9} \approx 2.04 \]
- Take the square root of both sides:
\[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
Since the time \( t \) when the rock hits the water is approximately 1.43 seconds, a reasonable scale for the range would be from 0 to slightly above 1.43.
Among the provided response options, the best scale for the range in terms of time would likely be 2, as it comfortably covers the time the rock reaches the water.
So, the answer is:
2