A rock is dropped into a well. The sound of the splash is heard 2.40s after the rock is released. How far below the top of the well is the surface of the water? If the travel time for sound is neglected, what percent error does this introduce to the depth of the well?

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A rock is dropped into a well. The sound of the splash is heard 2.40s after the rock is released. How far below the top of the well is the surface of the water? If the travel time for sound is neglected, what percent error does this introduce to the depth of the well?

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Let he depth be d. Then the time taken for the rock to reach d is given by the equation \[ d = \frac{1}{2}gt_1^2 \] then the time for the sound to return up the well is given by \[ v = d/t_2 \] where \( v \) is the speed of sound. You also know that \[ t_1 + t_2 = 2.6 \ sec \] Now use those three equations to find d.

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