fisica1-youn-e-freedman-exercicios-resolvidos

2.87: Take +y to be downward.
Last 1.0 s of fall:
1 2
2
y − y0 = v0
yt
+
2
a
yt
gives h 4 = v0
y
(1.0 s) + (4.9 m s )(1.0 s)
v 0y is his speed at the start of this time interval.
Motion from roof to y − y 0
= 3h
4 :
v
0
0, v = ?
y
=
y
2 2
2
v
y
= v0
y
+ 2a
y
( y − y0
)givesvy
= 2(9.80 m s )(3h
4) = 3.834 h m s
This is v y for the last 1.0 s of fall. Using this in the equation for the first 1.0 s gives
h 4 = 3.834 h + 4.9
2
Let h = u and solve for u : u = 16.5. Then h = u
2
= 270 m.
2
2.88: a) t + t 10.0 s
fall sound return
=
f
+ ts
=
t 10.0 s
(1)
d
Rock
1
gt
2
1
(9.8 m
2
= d
2
f
Sound
= v t
s
2
f
s
2
s ) t = (330 m s) t
(2)
Combine (1) and (2): t
f
= 8.84s,
ts
= 1.16s
m
h = vs ts
= (330 )(1.16s) = 383 m
s
b) You would think that the rock fell for 10 s, not 8.84 s, so you would have thought
it fell farther. Therefore your answer would be an overestimate of the cliff's height.
s
2.89: a) Let +y be upward.
2
y − y = −15.0 m, t = 3.25 s, a y
= −9.80
m s , v0
0 y
=
1 2
y − y0 = v0
yt
+
2
a
yt
givesv
0 y
= 11.31 m s
Use this v 0y in v v0 + a t to solve for v : v = −20.5
m
y
=
y y
y y
?
s
b) Find the maximum height of the can, above the point where it falls from the
scaffolding:
2
v = , v = + 11.31 m s, a = −9.80
m s , y − y ?
y
0
0 y
y
0
=
2 2
v
y
= v0
y
+ 2a
y
( y − y0
) gives y − y0
= 6.53 m
The can will pass the location of the other painter. Yes, he gets a chance.