A basketball of mass 0.60 kg is dropped from rest from a height of 1.05 m. It rebounds to a height of 0.57 m. (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of 1.05 m by exerting a constant downward force on it for a distance of 0.080 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 m, what is the magnitude of the force?

Chapter 4 Two and Three Dimensional Motion The best way to work with multidimensional vectors is by using component form The position is written as:r=xi+yj+zk In this case x, y, and z would represent constants Displacement is shown as: ∆ r=∆ xi+∆ yj+∆zk Average Velocity: v= ∆r ∆ t If needed break this up into component for and do the final minus the initial positon and divide it by time. Think of it as being the slope for the position (rise over run). Instantaneous Velocity: v= dr dt Acceleration: a= ∆v