A classmate released a stone at a certain time at a wellhead, and released a second stone after1s. After 4S, the first stone fell to the bottom of the well regardless of air resistance.

(1) The stone falls freely. According to the displacement time relation formula, there are:

h = 12gt 2 = 12× 10×42 = 80m

The speed at which the stone reaches the bottom of the well:

V = gt =10× 4 = 40m/s.

(2) The speed of the first stone is 30m/s, and the time is:

t 1 = v 1g = 30 10 = 3s

At this time, the displacement of the first stone is:

h 1 = 12gt 2 1 = 12× 10×9 = 45m

The displacement of the second stone is:

H2 = 12gt 22 = 12× 10×22 = 20m

So the distance between two stones is:

△h=45-20=25m

Answer: (1) The depth of this well is 80 meters; The speed of stones reaching the bottom of the well is 40m/s;

(2) When the speed of the first stone is 30m/s, the distance between two stones is 25m. ..