Question

While hiking, Julia kicked a small rock off a ledge that was meters above a crevasse. Use GeoGebra to graph the situation using the formula y=−4.9t2+9
. Use the graph to determine when the rock will hit the bottom of the crevasse, 2 meters below ground level. Round your answer to the nearest tenth of a second.

Answer 1

To analyze the motion of the rock using the equation \( y = -4.9t^2 + 9 \), we first need to understand what the variables represent. In this equation, \( y \) represents the height of the rock above the ground (in meters), and \( t \) represents time (in seconds). The rock is initially at a height of 9 meters, and the \( -4.9t^2 \) term represents the effect of gravity on the rock's motion.

The problem states that we want to find out when the rock reaches a height of 2 meters below ground level. Since ground level is considered 0 meters, 2 meters below ground level is represented as \( y = -2 \).

We need to set the equation equal to -2 and solve for \( t \):

\[
-4.9t^2 + 9 = -2
\]

Now, we can rearrange the equation to find \( t \):

\[
-4.9t^2 + 9 + 2 = 0
\]
\[
-4.9t^2 + 11 = 0
\]
\[
-4.9t^2 = -11
\]
\[
4.9t^2 = 11
\]
\[
t^2 = \frac{11}{4.9}
\]
\[
t^2 \approx 2.2449
\]
\[
t \approx \sqrt{2.2449} \approx 1.5
\]

Rounding this to the nearest tenth of a second, we find that \( t \approx 1.5 \) seconds.

### Summary
Using the graph of \( y = -4.9t^2 + 9 \), the rock will hit the bottom of the crevasse (2 meters below ground level) approximately **1.5 seconds** after it is kicked off the ledge.