# Water bucket in free fall

There is a situation that a bucket which is full of water and has a hole at the side is let to fall freely. Will the water flow out of the bucket? And if it does, will its path be parabolic or linear? Assume that there is no air and hydrodynamics is working. Full detailed answer would be appreciated.

I think water will not come out as evident by the Bernoulli's equation
p1 + dgh1 + 1\2dv1^2 = p2 + dgh2 +1\2dv2^2
here p1=p2= p atmospheric
dgh term will be zero as the system is in free fall
v1= 0

When the cup is at rest, the force of gravity pulls downward upon the water. At the location of the hole, there is nothing to balance gravity's force and prevent water from pouring out of the cup. However, when the cup is in free fall, the water will not leak. It is merely falling to the ground at the same rate as its surroundings (the cup).

berkeman
Mentor
There is a situation that a bucket which is full of water and has a hole at the side is let to fall freely. Will the water flow out of the bucket? And if it does, will its path be parabolic or linear? Assume that there is no air and hydrodynamics is working. Full detailed answer would be appreciated.

Welcome to the PF.

In the future, please post schoolwork-type questions in the Homework Help forums, and be sure to share your thoughts on how you think the problem can be solved. I have moved your thread to the HH forums.

Gravitational force disappears in a free-falling frame of reference, and in the absence of gravitational force, only one force remain: the cohesive force of attraction between water molecules. Hence, the mass of water will tend to acquire a spherical form. In so doing, constrained by the cylindrical or conical walls of the bucket, and its flat bottom, some water might be 'extruded' through the hole... For a stationary, non free-falling observer, the trajectory of the 'extruded jet' would be parabolic. For a free-falling observer, the extruded jet would move in straight line...

haruspex