114. We neglect air resistance, which justifies setting
a
= –
g
= –9.8 m/s
2
(taking
down
as
the –
y
direction) for the duration of the motion of the shot ball. We are allowed to use
Table 21 (with
Δ
y
replacing
Δ
x
) because the ball has constant acceleration motion. We
use primed variables (except
t
) with the constantvelocity elevator (so
'1
0
m
/
s
v
=
), and
unprimed variables with the ball (with initial velocity
0
20
30 m/s
vv
′
=+ =
, relative to the
ground). SI units are used throughout.
(a) Taking the time to be zero at the instant the ball is shot, we compute its maximum
height
y
(relative to the ground) with
22
00
2(
)
g
y
y
=−
−
, where the highest point is
characterized by
v
= 0. Thus,
yy
v
g
=+
=
o
m
0
2
2
76
where
oo
23
0
m
′
=
(where
o
28 m
y
′ =
is given in the problem) and
v
0
= 30 m/s
relative to the ground as noted above.
(b) There are a variety of approaches to this question. One is to continue working in the
frame of reference adopted in part (a) (which treats the ground as motionless and “fixes”
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Resistance, Acceleration

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