To determine when the rock hits the water, we need to find the time \( t \) when the height \( h \) is equal to 0 (i.e., when the rock reaches the lake). The equation given is:
\[ h = -4.9t^2 + 10 \]
Setting \( h \) to 0, we get:
\[ 0 = -4.9t^2 + 10 \]
Solving for \( t \):
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \approx 2.0408 \]
\[ t \approx \sqrt{2.0408} \approx 1.43 \]
Now that we've determined that the rock will hit the water at approximately \( t = 1.43 \) seconds, a reasonable scale for the range of the table to show this would be around the time intervals surrounding 1.43 seconds.
Considering typical time interval choices, a scale of 0.5 seconds or 1 second would be practical. Since the options provided include whole numbers like 2, 5, and the precise scale of 1 tenth is not reasonably practical, the appropriate answer would be 2 seconds, ensuring you capture the relevant time leading up to when the rock hits the water.
So, the answer is: 2.