How to play Leaky Tank
- Choose a test and read the hole's depth, its height, and its area on the card.
- Bernoulli's principle from the still surface to the hole turns pressure into speed: v = √(2·g·h).
- The jet then falls freely from the hole's height: x = v·√(2·y ÷ g). Type your prediction and press Start.
- Watch the jet land, and the chart on the right: deeper holes are faster, but lower holes have less time to fly.
How fast does water leave a tank?
Bernoulli's principle says that along a flow, pressure, motion, and height trade with one another: p + ½·ρ·v² + ρ·g·z stays the same. At the top of a tank the water is still and at air pressure; at a hole h lower the water is at air pressure again, so all the height it lost has become motion: ½·ρ·v² = ρ·g·h, or v = √(2·g·h). That is Torricelli's law, and it means water leaves a hole exactly as fast as a stone dropped from the surface would be moving after falling h. The deeper the hole, the faster the jet. Where the jet lands depends on its speed and on the time it takes to fall: x = v·√(2·y ÷ g). For a tank standing on the floor with water H deep, that becomes x = 2·√(h·(H − h)): a hole near the top is slow, a hole near the bottom is fast but low, and the hole halfway down shoots farthest. Two holes the same distance above and below the middle land together. The flow is Q = A·v, so a bigger or deeper hole empties a tank faster.
A focused learning model
A hose keeps the water level steady, the water is still at the surface, and the holes are small and sharp-edged with no losses, so v = √(2·g·h). Jets leave horizontally and fall without air resistance. A run lasts 2 s; for the bucket the hole stays open for 4 s and the run ends when the last water lands. Everything is to scale. Answers within 2% are right. g = 9.8 m/s². Plays in real time.