# Two drops come off the eaves with an interval of 1 s after how long will the distance between them be 25 m?

Let us denote the time in flight of the last detached drop by t. Then the time in flight of the drop that came off first will be (t + 1) s.

We find the paths traversed by the first and second drops, after time t, after the separation of the second drop:

s1 = (g (t + 1) ^ 2) / 2;

s2 = gt ^ 2/2.

The difference should be 25 m:

g (t + 1) ^ 2/2 – gt ^ 2/2 = 25;

g (t ^ 2 + 2t + 1) / 2 – gt ^ 2/2 = 25;

gt ^ 2 + 2gt + g – gt ^ 2 = 50;

2gt + g = 50;

(50 – g) / 2g = 40/20 = 2 s.

Answer. The first drop will be at a distance of 25 m from the second drop in 2 s, counting from the moment of separation of the second drop (or three seconds after the separation of the first

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