homework 04 – BAUTISTA, ALDO – Due: Jan 30 2006, 4:00 am
1
Version number
encoded
for clicker
entry:
V1:1, V2:4, V3:2, V4:4, V5:3.
Question 1
Part 1 of 2.
10 points.
A ball is thrown upward.
Its initial verti
cal speed is 8
.
6 m
/
s and the acceleration of
gravity is 9
.
8 m
/
s
2
,
as shown in the Fgure
below.
Neglect:
Air resistance.
8
.
6m
/
s
9
8m
2
h
max
What is its maximum height,
h
max
?
Correct answer:
3
.
77347
m (tolerance
±
1
%).
Explanation:
Let :
v
0
= 8
.
6 m
/
s
and
g
= 9
.
8 m
/
s
2
.
Basic Concept:
±or constant accelera
tion, we have
v
2
=
v
2
0
+ 2
a
(
x

x
0
)
.
(1)
Solution:
The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing
v
,
a
,
y
0
,
and
y
(no time) is
the easiest one to use.
Choose the positive
direction to be up; then
a
=

g
and
0 =
v
2
0
+ 2 (

g
) (
h
max

0)
or
h
max
=
v
2
0
2
g
=
(8
.
6 m
/
s)
2
2 (9
.
8 m
/
s
2
)
= 3
.
77347 m
.
Question 2
Part 2 of 2.
10 points.
±ind the speed
v
A
of the ball as the ball
passes a point
A
, which is at one quarter of
the maximum height
h
max
4
.
Correct answer: 7
.
44782
m
/
s (tolerance
±
1
%).
Explanation:
At the point
A
, the velocity is again found
from the same equation, but with the Fnal
velocity
v
A
v
2
A
=
v
2
0
+ 2 (

g
)
h
4
h
=
v
2
0
2
g
from Part 1, so
h
4
=
v
2
0
8
g
and
v
2
A
=
v
2
0

2
g
±
v
2
0
8
g
²
=
v
2
0

v
2
0
4
=
3
4
v
2
0
.
Thus
v
A
=
√
3
2
v
0
=
√
3
2
(8
.
6 m
/
s)
= 7
.
44782 m
/
s
.
Question 3
Part 1 of 2.
10 points.
An airplane ²ying parallel to the ground
undergoes
two
consecutive
displacements.
The Frst is 73 km at 27
.
9
◦
west of north,
and the second is 160 km at 69
.
3
◦
east of
north.
a) What is the magnitude of the plane’s
total displacement?
Correct answer: 167
.
336
km (tolerance
±
1
%).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenthomework 04 – BAUTISTA, ALDO – Due: Jan 30 2006, 4:00 am
2
Explanation:
73
km
117
.
9
◦
160
20
.
7
◦
d
Note:
Figure is not drawn to scale.
Basic Concepts:
Δ
x
=
d
(sin
θ
)
Δ
y
=
d
(cos
θ
)
Δ
x
total
= Δ
x
1
+ Δ
x
2
Δ
y
total
= Δ
y
1
+ Δ
y
2
d
total
=
q
(Δ
x
total
)
2
+ (Δ
y
total
)
2
Given:
d
1
= 73 km
θ
1
= 90
.
0
◦
+ 27
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Acceleration, Gravity, Work, – BAUTISTA

Click to edit the document details