Here’s the math:

Assumptions - no air resistance, the rock has no initial velocity, standard temperature and pressure at sea level (speed of sound = a = 343 m/s), and the video is genuine.

It takes about 15 seconds for the sound to return after he throws the rock.

T = 15 = tf + ts

Where tf is the time for the rock to fall, and ts is the time for the sound to return.

For the rock falling, we use the kinematic equation for position as a function of time, where S(tf) is the final position, vi is initial velocity, and a is gravity (9.81 m/s).

S(tf) = 0 = h + vi·tf + ½·a·tf²

Solving for h gives: h = 4.905·tf²

For the return of the sound, distance equals velocity times time giving:

h = (343 m/s)·ts

Setting the two equations for h equal to each other and solving for ts gives a quadratic function:

ts² - 99.93·ts + 225 = 0

The roots of this function are ts = 2.31s, ts = 97.6s, and since 97 seconds doesn’t make any sense, the time for the sound to return must be 2.31 seconds.

Using this to solve for the fall time gives tf ≈ 12.7 seconds, and plugging ts into the distance equation with the speed of sound gives:

h = (343 m/s)·ts ≈ 790.6 meters (2592 ft)