fisica1-youn-e-freedman-exercicios-resolvidos

3.25: Take + y to be downward.
a) Use the vertical motion of the rock to find the initial height.
2
t = .00 s, v = + 20.0 s, a = + 9.80 m/s , y − y ?
6
0 y
y
0
=
1 2
y − y0 = v0
y
t + a
yt
gives y − y0
= 296 m
2
b) In 6.00 s the balloon travels downward a distance y − y0 = (20.0 s)(6.00 s) = 120 m .
So, its height above ground when the rock hits is 296 m − 120 m = 176 m .
c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component
of the distance between the rock and the basket is 176 m, so the rock is
(176 m)
2
2
+ (90 m) = 198 m from the basket when it hits the ground.
d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0
m/s relative to the basket.
Just before the rock hits the ground, its vertical component of velocity is
2
v = v + a t = 20.0 s + (9.80 m/s )(6.00 s) 78.8 m/s , downward, relative to the ground.
y
0 y y
=
The basket is moving downward at 20.0 m/s, so relative to the basket the rock has
downward component of velocity 58.8 m/s.
e) horizontal: 15.0 m/s; vertical: 78.8 m/s
m
3.26: a) horizontal motion: x − x v
xt
t
60.0
0
=
0
so =
(v0
cos 43°
) t
vertical motion (take + y to be upward):
y − y
1 2
1
2
0
= v0
yt
+
2
a
yt
gives 25.0 m = ( v0
sin 43.0°
) t +
2
( −9.80 m/s )
Solving these two simultaneous equations for v
0
and t gives v = 3.26
0
m/s and t = 2.51s
.
b) v
y
when shell reaches cliff:
2
v
y
= v0 y
+ a
yt
= (32.6 m/s) sin 43.0° − (9.80 m/s )(2.51s) = −2.4 m/s
The shell is traveling downward when it reaches the cliff, so it lands right at the edge of
the cliff.
t
2
3.27: Take + y to be upward.
Use the vertical motion to find the time it takes the suitcase to reach the ground:
2
v0 y
= v0
sin 23°
, a
y
= −9.80 m/s , y − y0
= −114 m, t = ?
1 2
y − y0 = v0
y
t + a
yt
gives t = 9.60 s
2
The distance the suitcase travels horizontally is
x
x
− x 0
= v 0
= ( v 0
cos 23.0°
) t =
795 m