Final answer:
The velocity of a person falling from a 10 m high platform 2.21 m above water can be found using the conservation of energy. After calculation, the speed is approximately 11.26 m/s, which does not match the provided options.
Step-by-step explanation:
To determine the speed of a person falling from a platform before reaching water, we use the principles of kinematics. The energy conservation equation states that the sum of potential energy and kinetic energy is constant for a falling body in the absence of air resistance. When the person jumps from a platform, their potential energy is converted to kinetic energy as they fall. The potential energy (PE) at a height (h) is given by PE = mgh where m is the mass of the person, g is the acceleration due to gravity (9.81 m/s²), and h is the height above the ground. The kinetic energy (KE) of a moving object is given by KE = ½mv², where v is the velocity. At the moment the person is 2.21 m above the water, we know their potential energy (at that height) and we can equate it to her kinetic energy to find out her speed.
Let's calculate the initial potential energy at the 10 m platform and the potential energy at 2.21 m:
- Initial PE at 10 m: PE = mgh = 56 kg × 9.81 m/s² × 10 m
- PE at 2.21 m: PE = mgh = 56 kg × 9.81 m/s² × 2.21 m
Now, we calculate the kinetic energy at 2.21 m using the conservation of energy:
Initial PE at 10 m - PE at 2.21 m = KE at 2.21 m
KE at 2.21 m = ½mv²
By solving this equation, we can find the speed v.
After completing the calculations, we arrive at the answer that the person's velocity at 2.21 m above the water is approximately 11.26 m/s. However, none of the provided options exactly match this calculated value.