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()
00
ˆˆ
cos
i
sin
j.
vv
v
g
t
θθ
=+
−
G
(a) With
v
0
= 30 m/s and
θ
0
= 60°, we obtain
(15i+6.4 j) m/s
v
=
G
, for
t
= 2.0 s. The
magnitude of
v
G
is
22
 
(15 m/s)
(6.4 m/s)
16 m/s.
v
=
G
(b) The direction of
v
G
is
1
tan [(6.4m/s)/(15m/s)] 23 ,
−
==
°
measured counterclockwise from +
x
.
(c) Since the angle is positive, it is above the horizontal.
(d) With
t
= 5.0 s, we find
(15i
23j) m/s
v
=−
G
, which yields
 
(15m/s)
( 23m/s)
27 m/s.
v
−
=
G
(e) The direction of
v
G
is
1
tan [( 23m/s)/(15m/s)]
57
−
=
−
°
, or 57
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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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