Question

While hiking, Marek throws a rock off a cliff that is roughly 10 meters above a lake. This situation is modeled by the equation h=−4.9t2+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 hits the water, what would be a reasonable scale for the range?(1 point)
Responses

5
5

110
1 tenth

10
10

2

Answer 1

To determine 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 equal to 0 (when the rock reaches the water level).

Setting the height \( h \) to 0, we have:

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

Rearranging gives:

\[
4.9t^2 = 10
\]

Dividing by 4.9:

\[
t^2 = \frac{10}{4.9} \approx 2.04
\]

Taking the square root:

\[
t \approx \sqrt{2.04} \approx 1.43 \text{ seconds}
\]

To create a table for this scenario, we might want to include time values around 1.43 seconds. A reasonable scale for the range could be in increments of 0.5 seconds or 1 second to ensure we capture the moment just before and when the rock hits the water.

Among the provided options, the best choice for a scale that can represent the time from 0 to a little over 1.43 seconds is **1 tenth** (0.1 seconds). This allows for a detailed range of values leading up to the impact with the water.

So, the best answer is:

**1 tenth**