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which in turn gives
2
net,
/
(344.4 N)/(7.260 kg)
47.44 m/s .
xx
aF
m
==
=
Using Eq. 216 for constantacceleration motion, the speed
of the shot at the end of the acceleration phase is
22
2
0
2
(2.500 m/s)
2(47.44 m/s )(1.650 m)
12.76 m/s.
x
vv
a
x
=+
Δ
=
+
=
(b) If
42 ,
θ
=°
then
2
net,
2
sin
380.0 N (7.260 kg)(9.80 m/s )sin 42.00
45.78 m/s ,
7.260 kg
x
x
F
Fm
g
a
mm
−−
°
=
=
and the final (launch) speed is
2
0
2
(2.500 m/s)
2(45.78 m/s )(1.650 m)
12.54 m/s.
x
a
x
Δ
=
+
=
(c) The decrease in launch speed when changing the angle from 30.00
°
to 42.00
°
is
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 Spring '08
 Any
 Physics, Acceleration, Force

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