Answer :
Final answer:
Using conservation of energy, the velocity of a 100 kg object freely falling from 100 m is found to be 44 m/s. This is derived by equating the initial potential energy to the final kinetic energy and solving for velocity.
Explanation:
To determine the velocity of a freely falling 100 kg object at an elevation of 100 m using conservation of energy, we apply the principle that the initial potential energy is equal to the kinetic energy just before the object hits the ground (ignoring air resistance). The potential energy (PE) at 100 m can be calculated using PE = mgh, where m is the mass (100 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (100 m). The kinetic energy (KE) just before impact is given by KE = (1/2)mv², where v is the velocity we want to find.
To find the velocity, we set PE equal to KE: mgh = (1/2)mv². After canceling the mass, the equation simplifies to gh = (1/2)v². Solving for v gives v = √(2gh). Substituting the values, we get v = √(2 * 9.8 m/s² * 100 m), which equals 44 m/s. Therefore, the correct answer is (a) 44 m/s.
