**Unformatted text preview: **b) While the ring is in free fall, the average acceleration is the constant acceleration due to gravity, 2 s / m 80 . 9 down. c) 2 2 1 gt t v y y y-+ = 2 2 ) s m 8 . 9 ( 2 1 ) s m (5.00 m . 12 t t-+ = Solve this quadratic as in part a) to obtain t = 2.156 s. d) ) m . 12 )( s m 8 . 9 ( 2 ) s m (5.00 ) ( 2 2 2 2 2--=--= y y g v v y y s m 1 . 16 = y v e)...

View
Full
Document