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Thus, the speed of the stone when
max
/2
yy
=
is
22
2
2
(21.4 m/s)
( 12.7 m/s)
24.9 m/s
xy
vv
v
=+
=
+
−
=
.
(c) The percentage difference is
24.9 m/s 21.4 m/s
0.163 16.3%
21.4 m/s
−
==
.
30. (a) Since the
y
component of the velocity of the stone at the top of its path is zero, its
speed is
00
cos
(28.0 m/s)cos40.0
21.4 m/s
x
v
v
v
θ
=
=
=
°
=
.
(b) Using the fact that
0
y
v
=
at the maximum height
max
y
, the amount of time it takes for
the stone to reach
max
y
is given by Eq. 423:
sin
0s
i
n
y
v
g
t
t
g
−
¡
=
.
Substituting the above expression into Eq. 422, we find the maximum height to be
2
2
0
0
max
0
0
0
0
sin
sin
sin
11
(s
i
n)
s
i
n
.
2
v
yv
t
g
t
v
g
gg
g
θθ
§·
=−
=
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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

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