homework 02 – ASH, BEN – Due: Jan 30 2008, 11:00 pm
1
Question 1, chap 2, sect 6.
part 1 of 1
10 points
A
stone
is
thrown
straight
up
ward
and
at
the
top
of
its
trajec
tory
its
velocity
is
momentarily
zero.
What is its acceleration at this point?
1.
9.8 m/s
2
up
2.
Zero
3.
9.8 m/s
2
down
correct
4.
Unable to determine
Explanation:
Basic Concepts:
The gravitational accel
eration near the surface of the earth is consid
ered constant, for all practical purposes. This
acceleration of 9.8 m/s
2
is pointing downward.
Solution:
To illustrate how it works, let’s
take, for example, an upward initial velocity
of 9.8 m/s. One second later the velocity will
be zero. One second after that it will be at

9
.
8 m/s. In other words, in each second the
velocity is decreased by 9.8 m/s.
Question 2, chap 2, sect 6.
part 1 of 1
10 points
The acceleration due to gravity on planet
X is one Ffth that on the surface of the earth.
If it takes 4
.
7 s for an object to fall a certain
distance from rest on earth, how long would
it take to fall the same distance on planet X?
Correct answer: 10
.
5095 s (tolerance
±
1 %).
Explanation:
Because the acceleration due to gravity is
uniform near the surface of both planets, the
distance an object falls in a time
t
is given by
x
fall
=
g
planet
t
2
2
where
g
planet
is the force of gravity near the
surface of the given planet. Since the distance
x
fall
is the same for both planets, then
x
fall
=
g
e
t
2
e
2
=
g
x
t
2
x
2
=
1
5
g
e
t
2
x
2
t
2
e
=
1
5
t
2
x
t
x
=
√
5
t
e
=
√
5(4
.
7 s)
= 10
.
5095 s
.
Question 3, chap 2, sect 6.
part 1 of 4
10 points
A mountain climber stands at the top of a
50.9 m cli± hanging over a calm pool of water.
The climber throws two stones vertically 1.7
s apart and observes that they cause a single
splash when they hit the water.
The Frst
stone has an initial velocity of +2.3 m/s.
a) What will the velocity of the Frst stone
be at the instant both stones hit the water?
Correct answer:

31
.
6851 m
/
s (tolerance
±
1 %).
Explanation:
Basic Concept:
v
2
f
=
v
2
i
+ 2
a
Δ
y
Given:
Δ
y
=

50
.
9 m
v
i,
1
= 2
.
3 m
/
s
Δ
t
1
= Δ
t
2
+ 1
.
7 s
a
=

9
.
81 m
/
s
2
Solution:
v
2
f,
1
= (2
.
3 m
/
s)
2
+ 2
(

9
.
81 m
/
s
2
)
(

50
.
9 m)
= 1003
.
95 m
2
/
s
2
v
f,
1
=
±
r
1003
.
95 m
2
/
s
2
=
±
31
.
6851 m
/
s
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View Full Documenthomework 02 – ASH, BEN – Due: Jan 30 2008, 11:00 pm
2
The velocity is

31
.
6851 m
/
s since the motion
is downward.
Question 4, chap 2, sect 6.
part 2 of 4
10 points
b) How long after the release of the Frst
stone will the two stones hit the water?
Correct answer: 3
.
46434 s (tolerance
±
1 %).
Explanation:
Basic Concept:
v
f
=
v
i
+
a
Δ
t
Solution:
Δ
t
1
=
v
f,
1

v
i,
1
a
=

31
.
6851 m
/
s

(2
.
3 m
/
s)

9
.
81 m
/
s
2
= 3
.
46434 s
Question 5, chap 2, sect 6.
part 3 of 4
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 Spring '08
 ERSKINE/TSOI
 Acceleration, Gravity, Work, Velocity, Correct Answer

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