A ball falls from the top of a building . through the air(air resistance is present) to the ground below. How does the KE just before hitting the ground compare to the PE at the top of the building?

Isn't KE and PE equal to each other in this situation? Since I believe energey is conserved.

A ball droped some distance gains 30 J of Ke. Don't ignore air resistance. How much GPE did the ball lose?

Wouldn't the ball lose 30 J of gravitational potential energy?

If you account for friction, the KE has to be less than the original PE.

On the second, the GPE had to be greater than the finalKE. You have to know more about the air resistance to put a number of GPE.

If you ignore air resistance, then you are correct.

In the situation described, the ball does not have equal amounts of kinetic energy (KE) and potential energy (PE) at any given point. As the ball falls, its potential energy decreases while its kinetic energy increases. This is due to the conversion of potential energy into kinetic energy as the ball accelerates due to gravity.

Just before hitting the ground, the ball's potential energy would be nearly zero, as it is very close to the ground. At the same time, the ball's kinetic energy would be at its maximum before impact.

Regarding the second question, if the ball gains 30 J of kinetic energy while dropping, it means that it loses an equal amount of potential energy. This loss of potential energy is equivalent to its change in gravitational potential energy (GPE). Therefore, the ball would lose 30 J of gravitational potential energy during its descent.

In the first scenario, when a ball falls from the top of a building with air resistance, the KE (kinetic energy) just before hitting the ground will be greater than the PE (potential energy) at the top of the building. This is because some of the PE is converted into other forms of energy like sound and heat due to air resistance. The law of conservation of energy still holds true in this situation, but not all of the potential energy is converted into kinetic energy because of the presence of air resistance. So, the KE is lower than it would be if there were no air resistance, and the remaining energy is lost as heat and sound.

To calculate the exact values of KE and PE in this scenario, you would need to know the specifics of the ball, the height it's dropped from, and other factors. You would also need to consider the effects of air resistance, which can complicate the calculation. Generally, the potential energy at the top of the building is given by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. The kinetic energy just before hitting the ground is given by the formula KE = (1/2)mv^2, where m is the mass of the ball and v is its velocity.

In the second scenario, where a ball gains 30 J of kinetic energy after being dropped, it does not lose an equal amount of gravitational potential energy. This is because air resistance plays a role in the conversion of energy. Some of the initial gravitational potential energy is lost as heat and sound due to air resistance, so the loss in GPE will be less than 30 J. The exact amount will depend on the specifics of the situation, including the mass of the ball, the distance it falls, and the effects of air resistance.