To determine when the rock hits the water, we need to find when the height \( h \) is equal to 0 meters (the height of the lake). The equation given is:
\[ h = -4.9t^2 + 10 \]
Setting \( h \) to 0, we can solve for \( t \):
\[ 0 = -4.9t^2 + 10 \]
Rearranging gives us:
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \]
\[ t^2 \approx 2.04 \]
Taking the square root:
\[ t \approx \sqrt{2.04} \]
\[ t \approx 1.43 , \text{seconds} \]
Since the rock hits the water after approximately 1.43 seconds, a reasonable scale for the range of time \( t \) can be from \( 0 \) to \( 2 \) seconds. The best option that fits this range would be:
2 seconds.
Thus, the answer is 2.