The child is swinging on a rope swing. At the maximum distance from the equilibrium position

The child is swinging on a rope swing. At the maximum distance from the equilibrium position, its center of mass rises by 80 cm. The maximum speed of the child’s movement is?

To calculate the maximum speed of the child’s movement, we apply the law of conservation of energy (we represent the rope swing as a mathematical pendulum): m * Vmax ^ 2/2 = m * g * hmax and Vmax = √ (g * hmax * 2).

Variables and constants: g – gravitational acceleration (g ≈ 10 m / s2); hmax is the maximum distance of the center of mass from the equilibrium position (hmax = 80 cm = 0.8 m).

Let’s calculate: Vmax = √ (g * hmax * 2) = √ (10 * 0.8 * 2) = 4 m / s.

Answer: The maximum speed of the child’s movement is 4 m / s.