To determine when the rock hits the water, we need to find the value of \( t \) when the height \( h \) is equal to 0 (the height of the lake). The equation modeling the height of the rock is given by:
\[ h = -4.9t^2 + 10 \]
Setting \( h = 0 \):
\[ 0 = -4.9t^2 + 10 \]
Rearranging the equation gives:
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \]
\[ t^2 \approx 2.0408 \]
Taking the square root of both sides to solve for \( t \):
\[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]
Now, to create a table to determine when the rock hits the water, we want a reasonable time scale for the range of \( t \). Since the rock hits the water approximately at 1.43 seconds, a range that is just slightly larger than this, say from 0 to 2 seconds, would be appropriate.
Among the provided options, "2" would be the most reasonable choice for the range since it gives a complete view of the rock's height as it falls until it hits the water at approximately 1.43 seconds.
Thus, the answer is:
2