This is the most interesting and creative problem I solved attending KAIST. I officially cite our professor who made this question. This was on the general relativity and cosmology final exam.

I thank professor Ewan D. Stewart for bringing up such an insightful question for us.

Professor Stewart. Such a Great Mind.

He asked during the exam, how synchronized clocks would measure time differences under gravity. The official statement of the problem was as below:

Before solving the question, we see that the thrown clock experiences both velocity time dilation and gravitational time dilation. Since moving clocks measure shorter time (i.e. clock ticks slower) and due to gravity it also measures a slightly longer time (i.e. ticks faster), we must calculate the compromise between these two effects, and compare with the time measured by the held clock.

Near the Earth we use the Schwarzschild metric

$$ d\tau^2=(1-\frac{2GM}{r})dt^2-(1-2GM)dr^2-r^2d\Omega^2 $$

And since the thrown clock experiences purely radial motion (it has minimal angular motion), the equation of motion is

$$ \frac{d\tau}{dt}=\sqrt{(\frac{1-2GM}{r})-(1-\frac{2GM}{r})^{-1}(\frac{dr}{dt})^2} $$

The newtonian solution (i.e. holds well near Earth) for vertically launched clock is

$$ r\simeq r_0+v_0t-\frac{1}{2}g_0t^2 $$

where the gravitational acceleration

$$ g_0=\frac{GM}{r_0^2}. $$

Where the thrown clock returns at the Newtonian time coordinate

$$ t=\frac{2v_0}{g_0}. $$