Thermodynamical and mechanical point of view What is the difference in thermodynamical and mechanical point of view in treating systems.
For example if I drop a stone from height h what is thermodynamical view and mechanical view over it.
 A: Equilibrium Thermodynamics deals with equilibrium properties of macroscopic systems and wich new equilibrium states are reached or reachable after modification of external conditions or internal constraints of such systems. It is based on the consequences of very general principles, with the necessity of obtaining elsewhere data for specific systems. For instance, Thermodynamics is able to ensure that a stable homogeneous system at equilibrium must have a positive isothermal compressibility but it is not able to provide a numerical estimate of it without feeding in the theory some additional external information (equation of state of a specific material). From the point of view of equilibrium thermodynamics, full information on a specific system is contained in any of the possible fundamental equations like internal energy es a function of entropy, volume and number of particles, Helmholtz free energy , Gibbs free energy, enthalpy, entropy as a function of energy, volume and number of particles etc.
Equilibrium Statistical Mechanics aims to obtain the fundamental equations for a specific systems starting from a model of the microscopic interactions among the microscopic degrees of freedom, exploiting the consequences of a large (huge) number of degrees of freedom and using statistical arguments. In a way you may think statistical mechanics as a black box using statistics the input information about a Hamiltonian into a thermodynamic fundamental equation.
Coming to your example of a stone dropped from a height h, this is not a clean example of what a thermodynamic system is. At least, if the focus is on the dynamics of the stone. The reason is that the description of its dynamics, even taking into account the presence of air, can usually be obtained with  very reduced set of mechanical degrees of freedom, and also because, the interesting dynamic process of fall is not an equilibrium state.
An better example would be the description of the equilibrium states of a monoatomic perfect gas. Within Thermodynamics, its equilibrium properties are encoded into the two equations
$$
\begin{eqnarray}
U&=&\frac32 N k_B T \\
PV&=& N k_B T
\end{eqnarray}
$$
which fully characterize the thermodynamic behavior of the gas, but should be considered as given from experiments (they cannot be proved by Thermodynamics). On the contrary, Statistical Mechancs is able to prove, starting from the Hamiltonian, that a macroscopic system of non interacting monoatomic molecules at equilibrium must be described by the previous equations.
A: I  am not suggesting that these are formal “definitions”, just my view.
I would describe thermodynamics as the study of energy transfer in the form of heat and work between a system and its surroundings in relation to the properties of substances.
I would describe mechanics as the study of energy and forces and their effects on systems of physical bodies.
Both apply the law of conservatism of energy, although the focus of thermodynamics is internal enegy of a system whereas in mechanics it is the external macroscopic energy of a system, I.e. the energy of macroscopic bodies with respect to an external frame of reference. There is also the second law of thermodynamics for which I am not aware of a mechanics analogy.
Besides the second law, In my view heat transfer is the main difference between the two, being primarily in the realm of thermodynamics.
Regarding your example it deals with the conservation of macroscopic mechanical potential and kinetic energy. The stone originally has gravitational potential energy of $mgh$. While falling that energy is converted to kinetic energy. The sum of the potential and kinetic energy of the stone, neglecting air friction, is constant at any point in the fall. From a thermodynamics perspective that external energy of the system is part of the total energy of the system, the other part being its internal energy. The general form of the first law being 
$$\Delta E=\Delta U + \Delta KE + \Delta PE$$
Where $\Delta E$ is the total change in enegy of the system, $\Delta U$ Is the change in internal energy and $\Delta KE$ and $\Delta PE$ are the changes in kinetic and potential energy of the system as a. whole with respect to an external frame of reference.
Your example has no internal energy counterpart. For example it doesn’t deal with the temperature of the falling object or many other thermodynamic properties. Therefore, in my opinion, the example would not be the subject of thermodynamics.
Hope this helps.
