Potential and kinetic energy of equal objects who have reached terminal velocity from different heights

Question

My professor said that the potential energy of an object is larger the higher off the ground it is when it falls. I asked if there is a limit to that potential energy and she said no. I then asked about terminal velocity. Surely if two completely equal objects are dropped from heights sufficiently high enough for both of them to have reached terminal velocity, the height no longer matters pertaining to the amount of energy released? Whether one was up a mile further than the other, at some point they are both falling at the same rate and thus their potential energy is equal from that point on? In my mind, I then equate it to two equal objects falling at the same speed. They will both impact the ground with the same force, no? Where am I going wrong with this?

Edit: I appreciate the answers, but being a neophyte, I’m still slightly confused about the ultimate answer. My new understanding given the answers provided, is that the energy released by the object that has spent a longer time falling will be greater because it has had more energy gathered in some way due to having more time to heat up the air around it, and thus will actually impact with a greater energy released than the object with less time to fall. Is that correct?

Answer 1

Since you are talking about terminal velocity, you are including air resistance in this thought experiment. This means that there is another place for the potential energy of object to go. When an object reaches terminal velocity, it means that every bit of work done by gravity ($mg\Delta h$), goes into heating up the object and the air around it instead of increasing the velocity of the object. So, an object that falls from a greater height will reach the same terminal velocity as one that falls from a lesser height, but it will be hotter and have a greater amount of hotter air above it.

More technically, at terminal velocity, the decreasing potential energy of the object is accompanied by an increase in internal (thermal) energy of the object and air, so energy is conserved. The increase in internal energy is evidenced by the increased temperature of the object and the air it passes through

Answer 2

@MarkH already answared, but I want to add, that there indeed is a limit to potential energy, but for different reasons than you are suggesting.

For small heights the formula for gravitational potential energy is given by $$E_p=mgh,$$ where $m$ is mass of the object, $g$ is gravitational acceleration and $h$ is the height from the ground. This formula is of course unbounded, but it works only for small values of $h$. For larger values, you can no longer consider $g$ to be constant and you must take its dependence on $h$ into an account (the further you are from Earth, the weaker the gravity is). The correct formula for $h$ that varies a lot is: $$E_p=-G\frac{Mm}{h+r_e},$$ where $G$ is gravitational constant, $M$ mass of the Earth and $r_e$ is Earths radius.

For $h$ small compared to $r_e$ you get from this formula: $$E_p\approx-G\frac{Mm}{r_e}\left(1-\frac{h}{r_e}\right)=\text{const}+mgh,$$ where $g=G\frac{M}{r^2_e},$ which is the reason why for small heights the first formula is used. The constant term is irrelevant, because for computations only differences in energy matter. But it is negative, so the potential energy is negative number and as you go higher and higher this negative number gets smaller and smaller. In the limit of infinite height, it becomes zero, so the potential energy is actually bounded.

P.S. The approximation is given by Taylor expansion, which tells you that for $x \ll 1$ it holds $(1+x)^{-1}\approx 1-x.$ You can try it on calculator for small values of x.