fisica1-youn-e-freedman-exercicios-resolvidos

2.82: a)
2.83: a) From Eq. (2.14), with v 0 =0,
v
y
2
= 2a
( y − y0 ) = 2(45.0 m s )(0.640m) = 7.59m s.
y
v
b) The height above the release point is also found from Eq. (2.14), with
= 7.59 m s, v = 0and a = − ,
0 y
y
y
g
2
v0
y
h =
2 g
2
(7.59 m s)
=
= 2.94 m
2
2(9.80 m s )
2
45 m
(Note that this is also (64.0 cm) ( )
s
g
.The height above the ground is then 5.14 m.
c) See Problems 2.46 & 2.48 or Example 2.8: The shot moves a total distance 2.20 m
–1.83 m = 0.37 m, and the time is
(7.59 m s) +
(7.59 m s)
2
(9.80 m s
+ 2(9.80 m s
2
)
2
)(0.37 m)
= 1.60 s.