To determine when the rock hits the water, we need to find when the height \( h \) is equal to 0 (when it reaches the lake). The equation given is:
\[ h = -4.9t^2 + 10 \]
To find when the rock hits the water, we can set \( h = 0 \):
\[ 0 = -4.9t^2 + 10 \]
This can be rearranged to find \( t \):
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \]
\[ t^2 \approx 2.04 \]
Taking the square root of both sides:
\[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
Now, when setting up a table to determine when the rock hits the water, we should consider a reasonable time scale for \( t \), such as increments of seconds leading up to about 1.5 seconds.
Looking at the provided response options for the range:
- 2
- 10
- 5
- 110
- 1 tenth
A reasonable scale for the range would be 2, as it allows you to set up a table with entries like \( 0, 1, 2 \), which comfortably includes the time it takes for the rock to hit the water (approximately 1.43 seconds).