If a rock would hit the ground in 6.79 seconds, what would the height be

Can you please explain. Thanks.

This seems to fall in the realm of following equation
S = Ut + (1/2)at^2
Where S is the distance traveled, U is the initial velosicy, g is the acceleration and t is the time.
Assuming the initial velocity is 0 and only acceleration is the acceleration due to Earth's gravity the equation becomes

S = (1/2)(9.81)(6.79^2)
S = 226.14 meters

So the rock should have fallen from the height of 226.14 meters.

Height of fall = 1/2 (t^2 x g), = 225.91 metres.