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

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.<br /><br />Given:<br />- Initial height (h₀) = 5.0 m<br />- Buoyant force = 350% of her weight<br /><br />Let's denote the person's mass as m and the acceleration due to gravity as g (approximately 9.8 m/s²).<br /><br />The buoyant force is equal to 350% of her weight, which can be expressed as 3.5mg (since 100% of her weight is mg).<br /><br />The net force acting on the person is the difference between the buoyant force and her weight:<br />Net force = Buoyant force - Weight = 3.5mg - mg = 2.5mg<br /><br />According to Newton's second law, the acceleration (a) experienced by the person is:<br />a = Net force / m = 2.5g / m = 2.5g<br /><br />Now, we can use the kinematic equation to find the distance (H) the person goes down before reaching equilibrium:<br />H = h₀ - (initial velocity² / (2a))<br /><br />Since the person jumps feet first, we assume the initial velocity is negligible, so the equation simplifies to:<br />H = h₀ - (0² / (2a)) = h₀ - (0 / (2 * 2.5g)) = h₀ - (0 / 5g) = h₀<br /><br />Therefore, the person will go down to the initial height of 5.0 m before reaching equilibrium.<br /><br />So, the answer is:<br />H = 5.0 m