fisica1-youn-e-freedman-exercicios-resolvidos

2 249 m s
2.47: a) (224 m s) (0.9 s) = 249 m s . b) = 25.4.
c) The most direct way to
2
9.80 m s
ave
t = ((224 m s) 2)(0.9 s) =
find the distance is v
101m.
2
d) ( 283 m s) (1.40 s) = 202 m s but 40 g = 392 m s , so the figures are not consistent.
2
2
2.48: a) From Eq. (2.8), solving for t gives (40.0 m s – 20.0 m s )/9.80
b) Again from Eq. (2.8),
40.0 m s − ( −20.0
m s)
9.80 m s
2
=
6.12 s.
2
m s = 2.04 s.
c) The displacement will be zero when the ball has returned to its original vertical
position, with velocity opposite to the original velocity. From Eq. (2.8),
40 m s − ( −40
m s)
9.80 m s
(This ignores the t = 0 solution.)
d) Again from Eq. (2.8), (40 m s )/(9.80
time found in part (c).
2
e) 9.80 m s , down, in all cases.
f)
2
=
8.16 s.
2
m s ) = 4.08 s. This is, of course, half the