The body was thrown vertically upwards from a height of 40 meters and fell to the ground at a speed of 60 m / s

| February 15, 2021 | education

The body was thrown vertically upwards from a height of 40 meters and fell to the ground at a speed of 60 m / s. Determine the speed of the body at the moment of throwing. Disregard air resistance

To determine the speed of a given body at the moment of throwing, we will use the law of conservation of energy: Ekn + Enn = Ekk and m * Vn ^ 2/2 + m * g * hn = m * Vk ^ 2/2, whence we express: = Vk ^ 2/2 – g * h and Vn = √ (2 * (Vk ^ 2/2 – g * hn)).

Variables and constants: Vк – falling speed (Vк = 60 m / s); g – acceleration due to gravity (g ≈ 10 m / s2); hн – initial body height (hн = 40 m).

Let’s perform the calculation: Vn = √ (2 * (Vk ^ 2/2 – g * hn)) = √ (2 * (60 ^ 2/2 – 10 * 40)) ≈ 52.9 m / s.

Answer: At the moment of throwing, the given body acquired a speed of 52.9 m / s.



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