To determine when the rock hits the water, we need to find the time \( t \) when the height \( h \) becomes 0 (i.e., when the rock reaches the lake).
The equation given is:
\[ h = -4.9t^2 + 10 \]
Setting \( h = 0 \) (when the rock hits the water):
\[ 0 = -4.9t^2 + 10 \]
Solving for \( t \):
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \approx 2.04 \]
\[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
Since \( t \) is approximately 1.43 seconds, a reasonable scale for the range would be from 0 to around 2 seconds.
Given the options, here are the possible ranges:
- 10 seconds (too large)
- 2 seconds (reasonable)
- 5 seconds (too large)
- 110 seconds (much too large)
Therefore, the most appropriate choice for the scale of the range is 2 seconds.