fisica1-youn-e-freedman-exercicios-resolvidos

2.45: a) v
y
= v0 y
− gt = ( −6.00 m s) − (9.80 m s )(2.00 s) = −25.6
m s,
so the speed is
25 .6 m s .
b) 1 2
1
2
y = v
( 6.00 m s)(2.00 s) (9.80 m s )(2.00 s)
2
0 yt
− gt = −
−
= −31.6
m, with the
2
2
minus sign indicating that the balloon has indeed fallen.
c)
2 2
2
2
2 2
v v − 2g(
y − y)
= (6.00 m s) − 2(9.80 m s )( −10.0 m) = 232 m s ,so v = 15.2 m
y
=
0 y
0
y
2.46: a) The vertical distance from the initial position is given by
2
solving for v 0y ,
1 2
y = v0 yt
− gt
2
y 1 ( −50.0
m) 1
(9.80 m s
2
0
= + gt = +
)(5.00 s) =
t 2 (5.00 s) 2
v y
;
14.5 m s.
2 2
b) The above result could be used in v = v − g( y − y ), with v = 0, to solve for y
y
0 y
2
0
– y 0 = 10.7 m (this requires retention of two extra significant figures in the calculation
2
for v 0y ). c) 0 d) 9.8 m s , down.
e) Assume the top of the building is 50 m above the ground for purposes of graphing: