A stone is thrown vertically downwards from a height of 30 m above the ground with an initial speed

A stone is thrown vertically downwards from a height of 30 m above the ground with an initial speed of 5 m / s. What is the speed of the stone when it hits the ground?

To find the value of the speed of the specified stone at the moment of impact, we use the formula: S = h0 = (Vk2 – V0 ^ 2) / 2a = (Vk2 – V0 ^ 2) / 2g, whence we can express: Vk2 – V0 ^ 2 = h0 * 2g and Vk = √ (h0 * 2g + V0 ^ 2).

Constants and variables: h0 – initial height of the stone (h0 = 30 m); g – acceleration due to gravity (g ≈ 10 m / s2); V0 – throw speed, initial speed (V0 = 5 m / s).

Calculation: Vк = √ (h0 * 2g + V0 ^ 2) = √ (30 * (2 * 10) + 5 ^ 2) = 25 m / s.

Answer: At the moment of impact, the specified stone will have a speed of 25 m / s.