fisica1-youn-e-freedman-exercicios-resolvidos

2.43: a) Using the method of Example 2.8, the time the ring is in the air is
v
t =
0 y
+
v
2
0 y
− 2g(
y − y
g
2
(5.00 m s) + (5.00 m s) − 2(9.80 m s )( −12.0 m)
=
2
(9.80 m s )
= 2.156s,
12 .0 m
keeping an extra significant figure. The average velocity is then
2.156 s
= 5.57 m s , down.
As an alternative to using the quadratic formula, the speed of the ring when it hits the
2 2
ground may be obtained from v = v − g(
y − y ) , and the average velocity found
y
0 y
0
)
2
0
v y + v
from
0 y
2
; this is algebraically identical to the result obtained by the quadratic formula.
b) While the ring is in free fall, the average acceleration is the constant acceleration due
2
to gravity, 9.80m /s down.
2
1 2
c)
y = y
0
+ v
0 y t − gt
2
1
0 = 12.0m + (5.00m s) t − (9.8m s 2 ) t
2
Solve this quadratic as in part a) to obtain t = 2.156 s.
2 2
2
2
d) v = v − 2g(
y − y ) = (5.00 m s) − 2(9.8 m s )( −12.0 m)
e)
y
v
y
0 y
=16.1 m s
0
2