You are in a boat in a pool with a rock in your hand. You throw the rock into the pool. Does the water level rise, drop, or stay the same?

If the rock were neutrally buoyant the water level would remain the same. It is heavier than water which causes it to displace more than its own volume while in the boat compared to at the bottom of the lake. Therefore the water level of the lake would go down.

These answers are troubling. The only correct answer so far is Ben. The water level goes down.

The weight of the boat plus you plus the rock has already displaced the height of the water. The only time the water level will change will be when the rock is mid air.

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It will raise by the same volume of the rock.

First time, the water buoyant force is equal to the mass(boat + person + rock).g /subsequently, the water got displaced was also that much, the water level on the pool will be higher. second time, since rock is gonna deposit down to the bottom of the pool, it'll be like: water displacement = (boat + person).g + the volume of the rock (but it's probably much less than the water displacement it could have created when it's on the boat). There is no way that rock is displacing the same amount of water as it did previously. (1kg gold (Density: 19.30 g/cm3) and 1kg silver(Density: 10.49 g/cm3) in normal situation has different volume, so when deposited to the bottom, silver will displace more water than gold, yet both displace less water than when rock was sun-bathing on the boat) In conclusion, first time the water level was higher both on the pool wall and the mark on the boat, second time it'll be lower on both end.

Depends on how big the rock is

More weight on the boat = more displacement, while throwing the rock overboard would still displace water, the overall water lever will fall. Since the rocks surface area is smaller than the boat.

I'm not a mechanical engineer, but I did stay in a Holiday Inn Express last night. I hope no one else who has answered this question here (except for Homebrook) is a mechanical engineer. When you add the volume of the rock and subtract the volume of water previously displaced by the boat+rock, there is no change in the water level of the pool. Another way to answer is: The water level measured at the side of the pool remains the same. The boat becomes more buoyant and the water level measured at the side of boat falls.

you're already in the pool with the rock, so the water displaced is the same...this other guy is stupid

really? I thought the water would go up ?

At the moment the rock leaves your hand the water falls in the pool. When the rock enters the water the water level of the pool rises again to the level it had before you threw it.