To determine when the rock will hit the bottom of the crevasse using the equation \( y = -4.9t^2 + 9 \), we need to set \( y \) to -2 (since the bottom of the crevasse is 2 meters below ground level) and solve for \( t \).
The equation becomes:
\[ -4.9t^2 + 9 = -2 \]
Now, let's rearrange the equation:
\[ -4.9t^2 + 9 + 2 = 0 \] \[ -4.9t^2 + 11 = 0 \] \[ 4.9t^2 = 11 \] \[ t^2 = \frac{11}{4.9} \] \[ t^2 \approx 2.2449 \] \[ t \approx \sqrt{2.2449} \approx 1.5 \]
Now, rounding this to the nearest tenth of a second, we find that
The rock will hit the bottom of the crevasse in approximately \( 1.5 \) seconds.