homework 02 – ALIBHAI, ZAHID – Due: Jan 31 2007, 4:00 am
1
Question 1
part 1 of 2
10 points
The tallest volcano in the solar system is the
32 km tall Martian volcano, Olympus Mons.
Assume:
An astronaut drops a ball off the
rim of the crater and that the free fall acceler
ation of the ball remains constant throughout
the ball’s 32 km fall at a value of 2
.
7 m
/
s
2
.
(We assume that the crater is as deep as the
volcano is tall, which is not usually the case
in nature.)
a) Find the time for the ball to reach the
crater floor.
Correct answer: 153
.
96 s (tolerance
±
1 %).
Explanation:
Let :
h
= 32 km
and
a
= 2
.
7 m
/
s
2
.
Basic Concepts:
Free fall
s
=
s
o
+
v
0
t
+
1
2
a t
2
v
2
=
v
2
0
+ 2
a h .
Solution:
The distance an object falls
from rest under an acceleration
a
is
h
=
1
2
a t
2
,
so
t
=
radicalbigg
2
h
a
=
radicalBigg
2 (32 km)
(2
.
7 m
/
s
2
)
=
radicalBig
(172800 m
2
/
s
2
)
=
153
.
96 s
.
Question 2
part 2 of 2
10 points
b) Find the magnitude of the velocity with
which the ball hits the crater floor.
Correct answer: 415
.
692 m
/
s (tolerance
±
1
%).
Explanation:
Since the object falls from rest,
v
2
= 2
a h
so
v
=
√
2
a h
=
radicalBig
2 (2
.
7 m
/
s
2
) (32 km)
=
radicalBig
(172800 m
2
/
s
2
)
=
415
.
692 m
/
s
.
Question 3
part 1 of 2
10 points
A ball is thrown upward.
After reaching
a maximum height, it continues falling back
towards Earth.
On the way down, the ball
is caught at the same height at which it was
thrown upward.
Neglect:
Air resistance.
The acceleration
of gravity is 9
.
8 m
/
s
2
.
Its initial vertical speed
v
0
, acceleration
of gravity
g
, and maximum height
h
max
are
shown in the figure below.
v
0
9
.
8 m
/
s
2
h
max
If the time (
up
and
down
) the ball remains
in the air is
t
, calculate its speed
v
f
when it
caught.
1.
v
f
= 2
g t
2.
v
f
=
√
2
g t
3.
v
f
=
1
√
2
g t
4.
v
f
=
g t
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homework 02 – ALIBHAI, ZAHID – Due: Jan 31 2007, 4:00 am
2
5.
v
f
=
1
4
g t
6.
v
f
=
1
2
g t
correct
7.
v
f
= 4
g t
Explanation:
Basic Concept:
For constant accelera
tion, we have
v
0
=
g t
(1)
y
=
y
0
+
v
0
t
+
1
2
a t
2
.
(2)
Solution:
The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing
h
max
=
y
−
y
0
,
a
, and
t .
Choose
the positive direction to be up, then
a
=
−
g
and
t
up
=
v
0
g
.
From symmetry,
t
down
=
t
up
, we have
t
total
= 2
v
0
g
,
or
v
f
=
v
0
=
1
2
g t
total
.
Question 4
part 2 of 2
10 points
If the time the ball remains in the air is
t
,
calculate the maximum height
h
max
the ball
attained while in the air.
1.
h
max
= 4
g t
2
2.
h
max
=
1
4
g t
2
3.
h
max
= 8
g t
2
4.
h
max
=
1
8
g t
2
correct
5.
h
max
=
g t
2
6.
h
max
=
1
2
g t
2
7.
h
max
= 2
g t
2
Explanation:
The velocity at the top is zero.
Since we
know velocities and acceleration, Eq. 2 con
taining
h
max
=
y
−
y
0
,
a
, and
t .
Choose the
positive direction to be up, then
a
=
−
g
and
h
max
=
v
0
t
−
1
2
g t
2
=
g t
2
−
1
2
g t
2
=
1
2
g t
2
,
where
v
0
=
g t .
Therefore
t
up
=
radicalbigg
2
y
max
g
.
From symmetry,
t
down
=
t
up
, we have
t
total
= 2
radicalbigg
2
y
max
g
.
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 Spring '08
 Turner
 Acceleration, Work, Velocity, Correct Answer, Euclidean vector

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