Now let’s think of this from an energy perspective. The water is a closed system, so the total energy must be constant. There are two kinds of energy to think about in this case. First, there is gravitational potential energy (Ug = mgy). This is the energy associated with the height of an object above the surface of the Earth, and it depends on the height, the mass, and the gravitational field (g = 9.8 N/kg). The second type of energy is kinetic energy (K = (1/2)mv2). This is an energy that depends on the mass and the speed (v) of an object.

Since the total energy must be constant, the change in kinetic energy plus the change in gravitational potential energy must be equal to zero. The water at the top of the bucket (which I will call position 1) is stationary, and the water at the bottom (position 2) has some exit velocity, v. Putting this together, I get the following:

Notice that this is just the magnitude of the velocity. It actually doesn’t matter if this hole points straight down from the bottom of the bucket or horizontally out through its side—the water’s speed will be the same. But let’s say that the hole is on the side, so that the water shoots out parallel to the ground. If the distance from the hole to the ground is y0, how far from the hole will the water stream land when it hits the ground?

Even though this would be a stream of water, we can treat each molecule like an individual particle with just the downward-pulling gravitational force acting on it. Yes, this is your classic projectile motion problem.

The key idea here is that we can separate the horizontal and vertical motion into two separate problems. This means that in the vertical direction, the shooting water is the same as if it were just a single drop falling straight down, with an initial vertical velocity of 0 meters per second (m/s) since the water was shot horizontally. We can use this vertical motion to determine the time it takes to fall down to the ground. For the horizontal direction, it’s just a drop of water moving with a constant velocity—the same velocity that it was shot from the hole. Using the time from the vertical motion, I can calculate how far the water travels.