Rotational Kinetics Help!?

Question

A ball of radius 0.212m rolls along a horizontal table top with a constant linear speed of 5.51m/s. The ball rolls off the edge and falls a vertical distance of 2.17m before hitting the floor. What is the angular displacement of the ball while it is in the air?

(I am not using this for a homework question but am studying questions from a textbook. It would be great if I could learn how to approach this question, thanks!)

so I tried using this method and did not end up with the correct answer. Here's what I did:

C=(pie)d=1.26m

1.26/3.60m/s = 0.349rad/s

2.10 = 1/2(9.8)t^2 = 0.654s

0.349rad/s/0.654s = 0.534rad

The actual answer should be 11.8rad. What went wrong?

Answer 1

For Rads /sec, think radius/sec

5.51m/sec *1radius/.212m
= 26 Radius/sec
=26 rads/sec

Now time
t^2=2d/a
t^2=.443
t=.665 sec

Angular Displace = Rads/sec *t
=17.3 rads
(Not quite sure why I got a different number than your book).

Answer 2

Find circumference of the ball, then divide that by linear speed to get revolutions per second. Then find time falling by 2.17=1/2 g t^2. Divide revolutions/sec by time falling to get revolutions in the air.