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49. Using Eq. 426, the launch speed of the projectile is
2
0
(9.8 m/s )(69 m)
26.52 m/s
sin 2
sin 2(53 )
gR
v
θ
==
=
°
.
The horizontal and vertical components of the speed are
0
0
cos
(26.52 m/s)cos53
15.96 m/s
sin
(26.52 m/s)sin53
21.18 m/s.
x
y
vv
°
=
°
=
Since the acceleration is constant, we can use Eq. 216 to analyze the motion. The
component of the acceleration in the horizontal direction is
2
2
2
(15.96 m/s)
40.7 m/s ,
2
2(5.2 m)cos53
x
x
v
a
x
=
°
and the force component is
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 Spring '08
 Any
 Physics

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