A basketball player makes a jump shot. the 0.583 kg ball is released at a height of 1.80 m above the floor with a speed of 7.29 m/s. the ball goes through the net 3.10 m above the floor at a speed of 4.22 m/s. what is the work done on the ball by air resistance, a nonconservative force?

The kinetic and potential energies must be taken into account. The ball lost some speed but was also raised in the vertical direction between the two measured positions. This will cost some of the kinetic energy that we don't want to ascribe to the air resistance losses.

The gain in potential energy is

mg (h₂-h₁) = 0.583*9.81 * (3.10-1.80) = 7.435J

The loss in kinetic energy is

(1/2) m (v₁² - v₂²) = (0.583/2) (7.29²-4.22²) = 10.3J

Since 7.435J of this is due to the fact that it gained potential energy the final result is E=10.3J-7.435J=2.865J.

One way to think about the potential energy part is that it is higher in the air at the hoop than when the player releases the ball. In other words it hasn't fallen as far down as it went up, it is still in the process of gaining speed on the way down. This reduction in speed means less kinetic energy, but not because it was lost. It will be gained once it falls to the height of 1.8m, minus the little loss it will experience over that distance due to air resistance.