This situation is modeled by the equation h = -4.9 t2 + 10, with h representing the height in meters and t the time in seconds. If you were to set up a table to determine when the rock hit the water, what would be a reasonable scale for the range?

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
3 hours ago

To determine a reasonable scale for the range in a table that models when the rock hits the water using the equation \( h = -4.9t^2 + 10 \), we need to find the time \( t \) when the height \( h \) is zero (i.e., when the rock hits the water).

Setting \( h = 0 \):

\[ 0 = -4.9t^2 + 10 \]

Rearranging the equation gives:

\[ 4.9t^2 = 10 \]

Dividing both sides by 4.9:

\[ t^2 = \frac{10}{4.9} \approx 2.0408 \]

Now, taking the square root of both sides to solve for \( t \):

\[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]

Now that we know the rock hits the water at approximately \( t = 1.43 \) seconds, we can set up a table with a reasonable time scale. It is useful to include points before and after this time to see how the height changes.

A reasonable time scale might range from \( 0 \) to \( 2 \) seconds, with intervals of \( 0.2 \) or \( 0.5 \) seconds. Here’s a sample table with increments of \( 0.5 \) seconds:

| Time (t) | Height (h) = -4.9t² + 10 | |----------|---------------------------| | 0.0 | 10 | | 0.5 | 7.225 | | 1.0 | 5.0 | | 1.5 | 2.225 | | 2.0 | 0 |

This table consists of several key time points that illustrate the change in height until the rock reaches the water at \( t = 2.0 \) seconds. Note that since our calculated hit time is around \( 1.43 \) seconds, the value at \( t = 1.5 \) gives a good approximation of when the height is getting very close to zero.