With what time interval did two drops come off the cornice, if, 2 seconds

With what time interval did two drops come off the cornice, if, 2 seconds after the start of the fall of the second drop, the distance between them was 25 m.

In two seconds, drop # 2 flew distance S1. By the formula S1 = gt ^ 2/2, then:
S = 9.81 * 2 * 2/2 = 19.62 m.
Drop No. 1 flew S2 = 19.62 + 25 = 44.62 m.
The time of its movement will be equal to t = √ (2S / g):
t = √ ((2 * 44.62) / 9.81) = 3.016 s.
The difference is 3.016 – 2 = 1.016 s.
Answer: the difference is about 1 second.