Final answer:
The kinetic energy of the basketball when it reaches 0.5 m after being dropped is 4.9 J, calculated using the conservation of energy principle and the formula for potential energy.
Step-by-step explanation:
To determine the kinetic energy of the basketball when it reaches 0.5 m, we need to use the conservation of energy principle. The total mechanical energy (potential plus kinetic energy) of the basketball is conserved if we ignore air resistance and other non-conservative forces.
At the height of 1 m, the basketball has potential energy (PE) given by the equation PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is height. Since the ball is held and not moving, its initial kinetic energy (KE) is 0, so all the energy is potential. The potential energy at 1 m is:
PE1m = (1 kg)(9.8 m/s2)(1 m) = 9.8 J
When the ball drops to 0.5 m, it loses some potential energy and gains kinetic energy. The potential energy at 0.5 m is:
PE0.5m = (1 kg)(9.8 m/s2)(0.5 m) = 4.9 J
Since the total mechanical energy is conserved, the loss in potential energy is equal to the gain in kinetic energy. Therefore, the kinetic energy of the basketball at 0.5 m is the difference between its initial potential energy and its potential energy at 0.5 m:
KE = PE1m - PE0.5m = 9.8 J - 4.9 J = 4.9 J
Thus, the correct answer to the question 'What is the Kinetic Energy of the ball when it reached 0.5 m after it is dropped?' is (b) 4.9 J.