fisica1-youn-e-freedman-exercicios-resolvidos

5.106: (a) To find find the maximum height and time to the top without fluid resistance:
2 2
v = v + 2a(
y − y )
y
y − y
0
v
=
0 y
2
y
− v
2a
2
0 y
0
2
0 − (6.0 m s)
=
= 1.84 m or 1.8 m
2
2( −9.8 m s )
v − v0
0 − 6.0 m s
t = =
= 0.61s
2
a − 9.8 m s
(b) Starting from Newton’s Second Law for this situation
dv
m = mg − kv
dt
we rearrange and integrate, taking downward as positive as in the text and noting that the
velocity at the top of the rock’s “flight” is zero:
0
dv k
∫ = − t
v − v m
v
ln( v − v )
t
t
0
v
− vt
= ln
v − v
t
− 2.0 m s
= ln
= ln(0.25) = −1.386
− 6.0 m s − 2.0 m s
2
2
From Eq. 5.9, m k = v g = (2.0 m s ) (9.8 m s ) 0.204 s,
and
t
=
m
t = − k
( −1.386)
= (0.204 s) (1.386) = 0.283 s to the top. Equation 5.10 in the text gives us
dx
−(k
m) t
−(k
m) t
= vt
(1 − e ) = vt
− vte
dt
x t t
∫ dx = ∫v
dt − ∫v e
t
0 0 0
vtm
= vtt
+ ( e
k
= (2.0 m
= 0.26 m
−(k
m) t
s) (0.283
−(k
m) t
t
−1)
dt
s) + (2.0 m s) (0.204 s)(e
−1.387
−1)