To determine when the rock will hit the bottom of the crevasse (2 meters below ground level), we need to set the equation of the rock's motion \( f(x) = -4.9x^2 + 9 \) equal to -2 (since the bottom of the crevasse is 2 meters below ground level). This gives us the equation:
\[ -4.9x^2 + 9 = -2 \]
Now, let's solve for \( x \):
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First, rearrange the equation: \[ -4.9x^2 + 9 + 2 = 0 \] \[ -4.9x^2 + 11 = 0 \]
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Next, isolate the term involving \( x \): \[ -4.9x^2 = -11 \] \[ 4.9x^2 = 11 \]
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Divide both sides by 4.9: \[ x^2 = \frac{11}{4.9} \] \[ x^2 \approx 2.2449 \]
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Take the square root of both sides: \[ x \approx \sqrt{2.2449} \approx 1.5 \]
Thus, to the nearest tenth of a second, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.