Final answer:
The velocity of water exiting from the bottom of a water tower (vt) is equal to the velocity of water that free falls from the same height (vf) under ideal conditions without air resistance, as described by Torricelli's theorem and the kinematic equation for free-falling objects.
Step-by-step explanation:
When comparing the velocity of water exiting at the bottom of a water tower (vt) versus the velocity of free-falling water from the same height (vf), it is key to understand Torricelli's theorem and the concept of conservation of energy. According to Torricelli's theorem, if we ignore air resistance and assume that the fluid is incompressible and non-viscous, the speed of water exiting the tower only depends on the gravitational acceleration g and the height h of the water level above the outlet (v² = 2gh).
The same kinematic equation that is used to describe the motion of free-falling objects, v² = v² + 2gh, holds true for the water ejecting from the tower. Here, the initial velocity (v²) is zero, resulting in the final equation vt² = 2gh for water exiting the tower. In the case of free-falling water, using the same equation with initial velocity being zero, we get vf² = 2gh. This shows that the velocity of water exiting the water tower (vt) will be equal to the velocity of the water that has free-fallen from the same height (vf) under the ideal conditions as mentioned.