A ball of radius 0.2m rolls along a horizontal table top with a constant linear speed of 3.04 m/s. The ball rolls off the edge and falls a vertical distance of 2.02m before hitting the floor. What is the angular displacement of the ball while it is in the air.

Answer 1

Angular spin rate (w) remains the same after falling off the edge. Multiply w by the time it takes to fall 2.02 m.

w = V/R

Time to fall H = sqrt(2H/g)
= 0.642 s when H = 2.02 m