# The body, thrown vertically upward, has been at an altitude of 200 m twice with an interval of 6 seconds.

The body, thrown vertically upward, has been at an altitude of 200 m twice with an interval of 6 seconds. What is the total body flight time?

Task data: h1 (the height at which the body was taken twice) = 200 m; Δt (time interval between passes of a given height) = 6 s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) The duration of the rise (fall) from (to) a height of 200 m: t1 = t2 = Δt / 2 = 6/2 = 3 s.

2) Distance between maximum height and 200 m: h2 = g * t1 ^ 2/2 = 10 * 3 ^ 2/2 = 45 m.

3) Total flight time: t = 2 * √ (2h / g) = 2 * √ ((h1 + h2) / g) = 2 * √ (2 * (200 + 45) / 10) = 14 s.

Answer: The total flight time of the taken body is 14 seconds.

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