If one derives the Bernoulli Equation for the isentropic ideal gas flow you get:
$$ \frac{1}{2}v^2 + g z + \left(\frac{\kappa}{\kappa -1} \right) \frac{p}{\rho} = const. $$
Two questions:
Is isentropic equivalent to frictionless?
If the temperature stays the same during the process the quotient of pressure and density stays the same? This would lead to the very easy formula for the velocity at $t_2$. $$v_2=\sqrt{v_1^2-gz}$$
Yes, after a fashion, increase of entropy is always coupled to some kind of "friction". Though isentropic could be argued to be stronger than frictionless, as it does not only refer to mechanical friction. For example a Rayleigh flow with heat transfer will not be isentropic (credits to Azad for pointing out the example).
Nearly, it is $v_2 = \sqrt{v_1^2 - 2g \Delta z}$. But note, that constant temperature does not necessarily imply isentropicity.
