A person jumps from a height of 5.0 m into a swimming pool entering feet first with her hands at her sides. She maintains this position upon entering the water and experiences a constant upward force equal to 350% of her weight due to the water itself. How far down does she go? Express your answer with the appropriate units. H= Value Units

Question

A person jumps from a height of 5.0 m into a swimming pool entering feet first with her hands at her sides. She maintains this position upon entering the water and experiences a constant upward force equal to 350% of her weight due to the water itself. How far down does she go? Express your answer with the appropriate units. H= Value Units

Answer 1

To solve this problem, we need to use the principles of mechanics, specifically the concept of buoyant force and the equations of motion under constant acceleration.Given:- Initial height (h₀) = 5.0 m- Buoyant force = 350% of her weightLet's denote the person's mass as m and the acceleration due to gravity as g (approximately 9.8 m/s²).The buoyant force is equal to 350% of her weight, which can be expressed as 3.5mg (since 100% of her weight is mg).The net force acting on the person is the difference between the buoyant force and her weight:Net force = Buoyant force - Weight = 3.5mg - mg = 2.5mgAccording to Newton's second law, the acceleration (a) experienced by the person is:a = Net force / m = 2.5g / m = 2.5gNow, we can use the kinematic equation to find the distance (H) the person goes down before reaching equilibrium:H = h₀ - (initial velocity² / (2a))Since the person jumps feet first, we assume the initial velocity is negligible, so the equation simplifies to:H = h₀ - (0² / (2a)) = h₀ - (0 / (2 * 2.5g)) = h₀ - (0 / 5g) = h₀Therefore, the person will go down to the initial height of 5.0 m before reaching equilibrium.So, the answer is:H = 5.0 m

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