Drops are falling from the roof. Each subsequent drop breaks off from the roof 2 seconds after the previous one breaks off. How many seconds after the separation of the first drop, the distance between the first and third drops will be equal to 40 meters. neglect air resistance.
Let the first drop in t₁ seconds flew h₁ meters, and the third drop in t₂ seconds flew h₂ meters. From the condition of the problem, it is known that drops fall from the roof, and each subsequent drop breaks off from the roof 2 seconds after the separation of the previous one and after the separation of the first drop after some time t₁ = t₂ + 2 + 2 = t₂ + 4 (s), the distance between the first and the third drop was equal to 40 meters, that is, h₁ = h₂ + 40 (m). Since the air resistance can be neglected, then h₂ = g ∙ t₂² / 2 and h₁ = g ∙ t₁² / 2, where the proportionality coefficient g = 9.8 N / kg. We get, h₂ + 40 = g ∙ (t₂ + 4) ² / 2 or g ∙ t₂² / 2 + 40 = g ∙ (t₂ + 4) ² / 2; t₂ = 0.04 (s), t₁ = 0.04 + 4 = 4.04 (s).
Answer: in 4.04 seconds.
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