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63. We choose
down
as the +
y
direction and place the coordinate origin at the top of the
building (which has height
H
). During its fall, the ball passes (with velocity
v
1
) the top of
the window (which is at
y
1
) at time
t
1
, and passes the bottom (which is at
y
2
) at time
t
2
.
We are told
y
2
–
y
1
= 1.20 m and
t
2
–
t
1
= 0.125 s. Using Eq. 215 we have
yy
v
tt
g
211
2
1
2
1
2
1
2
−=
−+
−
bg bg
which immediately yields
()
2
2
1
2
1
1.20 m
9.8 m/s
0.125 s
8.99 m/s.
0.125 s
v
−
==
From this, Eq. 216 (with
v
0
= 0) reveals the value of
y
1
:
2
2
11
1
2
(8.99 m/s)
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics

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