The body is thrown vertically from the ground above the pit at a speed of 20 m / s.

The body is thrown vertically from the ground above the pit at a speed of 20 m / s. Where will the body be in 5 seconds? The depth of the pit is 30 m.

Given: V0 (throwing speed) = 20 m / s; t (considered travel time) = 5 s; h2 (pit depth) = 30 m.

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

1) Duration of vertical rise: t1 = V0 / g = 20/10 = 2 s.

2) Lift height: h1 = V0 * t / 2 = 20 * 2/2 = 20 m.

3) Remaining time for falling: Δt = t – t1 = 5 – 2 = 3 s.

4) Distance traveled during the fall: S = g * Δt ^ 2/2 = 10 * 3 ^ 2/2 = 45 m.

5) Location of the body after 5 s: Δh = h1 + h2 – S = 20 + 30 – 45 = 5 m.

Answer: After 5 s, the body will be at a depth of 25 m (5 m above the level of the pit).




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