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.