8.01T
Problem
Set
2
Solutions
Fall
2004
Problem
1.
Measurement
of
g
.
a)
The
ball
moves
in
the
vertical
direction
under
the
inﬂuence
of
the
con-
stant
force
of
gravity
1
.
Hence
in
our
approximation
the
ball
undergoes
one-
dimensional
motion
with
constant
acceleration
g
.
Let
us
introduce
vertical
axis
x
and
choose
the
positive
direction
to
point
up.
We
then
place
the
origin
of
the
coordinate
system
at
the
bottom
of
the
lower
pane
of
glass
and
choose
to
measure
the
time
from
the
moment
when
the
ball
rises
past
the
origin,
i.e.
we
choose
x
0
=
0
and
t
0
=
0
(See
Fig.
1).
With
our
choice
of
initial
position
Figure
1:
and
time,
the
equation
for
x
(
t
)
is
gt
2
x
(
t
)
=
v
0
t
−
(1)
2
and
the
graph
of
x
(
t
)
should
look
like
the
one
on
Fig.
1
The
velocity
of
the
ball
is
given
by
v
(
t
)
=
v
0
−
gt
(2)
b)
Using
Fig.
1,
it
is
easy
to
understand
that
the
time
the
ball
travels
across
the
window,
i.e.
the
interval
of
time
between
the
moment
it
rises
past
the
bottom
of
the
lower
plane
of
glass
and
disappears
past
the
top
of
the
upper
pane,
is
�
=
(
T
A
−
T
B
)
/
2.
Since
the
height
of
the
glass
is
h
,
we
obtain,
from
Eq.
(1)
g�
2
h
=
v
0
�
−
(3)
2
1
We
neglect
the
air
resistance
and
the
variation
of
the
gravitational
force
over
the
height
of
the
ball’s
trajectory
1

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8.01T
Problem
Set
2
Solutions
Fall
2004
To
find
the
second
equation,
we
notice
that
the
highest
point
of
the
trajectory
corresponds
to
v
=
0
and
t
=
T
A
/
2.
Thus
we
can
find
v
0
as
v
0
=
gT
A
/
2
(4)
We
now
substitute
this
result
back
into
Eq.
(3)
and
obtain
2
h
8
h
g
=
�
(
T
A
−
�
)
=
T
2
A
−
T
2
B
(5)
Problem
2.
Track
event.

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