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64. We first use Eq. 426 to solve for the launch speed of the shot:
2
0
2
(tan )
.
2( cos )
gx
yy
x
v
θ
−=
−
′
With
34.10 ,
=°
0
2.11 m
y
=
and
( , )
(15.90 m,0)
xy
=
, we find the launch speed to be
11.85 m/s.
v
′ =
During this phase, the acceleration is
22
2
0
(11.85 m/s)
(2.50 m/s)
40.63 m/s .
(
1
.
6
5
m
)
vv
a
L
′ −
−
==
=
Since the acceleration along the slanted path depends on only the force components along
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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, Acceleration

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