# A body thrown vertically upwards from a height of 40 m with an initial speed, the modulus

October 2, 2021 | education

| **A body thrown vertically upwards from a height of 40 m with an initial speed, the modulus of which is 5 m / s. At what height will the body be after a time interval of 2 s?**

To find the height at which a given body should be in 2 seconds, apply the formula: S = h = h0 + V0 * t + a * t ^ 2/2 = h0 + V0 * t – g * t ^ 2/2.

Constants and variables: h0 – initial height (h0 = 40 m); V0 is the modulus of the initial speed (V0 = 5 m / s); t – time interval (t = 2 s); g – acceleration due to gravity (g ≈ 10 m / s2).

Calculation: h = h0 + V0 * t – g * t ^ 2/2 = 40 + 5 * 2 – 10 * 2 ^ 2/2 = 40 + 10 – 20 = 30 m.

Answer: After 2 s, the given body will be at a height of 30 m.

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