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If a rock is dropped into awell that 340 feet deep how many seconds will it take for the rock to hit the bottom?
Wiki User
∙ 15y ago
This problem is difficult to answer without more information. Is the well filled with water? If so then you must know the size of the rock and the shape of the rock to calculate the affect of buoyancy and friction due to viscosity. If you assume the rock is dropping through air the only force is that of gravity. Therefore you can use the acceleration due to gravity and the following equation: D= Vinitial*t +(1/2)g*t^2 V initial = 0 here D=340 ft Than solving for time you get: t = 4.6 seconds However if you include all the forces you have the equation: (Mass of rock-density of fluid*volume of rock)*g -force viscosity = mass of rock*acceleration If you take the rock to be a sphere, stokes law under streamline conditions gives viscosity to be F=6*pi*viscosity*radius of sphere*velocity Plugging this in gives you a differential equation that can be solved however in order to solve this problem you would need to assume that the rock reaches terminal velocity before it hits the bottom. At this point the forces add up to zero and you can find a formulation for the terminal velocity and plug it into your differential equation. Sounds difficult but it is just simple mathematics.
Wiki User
∙ 15y ago
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