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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