To determine when the rock hits the water, we need to find out when the height \( h \) is 0 (i.e., when the rock reaches the level of the lake). The equation given is:
\[ h = -4.9t^2 + 10 \]
Setting \( h = 0 \) for when the rock hits the water gives us:
\[ 0 = -4.9t^2 + 10 \]
Rearranging this, we get:
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \]
\[ t^2 \approx 2.0408 \]
Taking the square root gives:
\[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]
Now that we know the rock will hit the water at approximately 1.43 seconds, we should consider a reasonable range for our table. Since it makes sense to have values that surround the time when the rock hits the water, a range extending to perhaps 2 seconds would be reasonable.
Looking at the options provided for scale:
A) 10
B) 2
C) 1/10
D) 5
The best option that allows us to comfortably accommodate our time of approximately 1.43 seconds, while also providing enough detail to see the change in height throughout the drop, would be B) 2. This lets us create a table with increments up to 2 seconds and captures the moment right before the rock hits the water effectively.