physics.....?

Question

A basketball player makes a jump shot. The 0.650-kg ball is released at a height of 1.90 m above the floor with a speed of 8.00 m/s. The ball goes through the net 3.10 m above the floor at a speed of 3.90 m/s. What is the work done on the ball by air resistance, a nonconservative force?
J

Answer 1

Hi Big T:

This is an energy problem.

Intitial Energy = PE + KE

PE= m*g*h=.65kg*9.8m/sec^2* 1.9m
KE=1/2 mv2= .5*.65kg*8.0^2
Initial energy = 32.9 Joules

Final Energy = .65 *9.8m/sec*3.1 +.5*.65*(3.9)^2

Air Resistance= Initial Energy - Final Energy

Your final answer should be ~ 8 Joules.