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(a) Placing the origin at the bottom of the plane, the kinematic equations (Table 21) for
motion along the
x
axis which we will use are
vv
a
x
2
0
2
2
=+
and
vv a
t
0
. The block
momentarily stops at its highest point, where
v
= 0; according to the second equation, this
occurs at time
tv
a
=−
0
. The position where it stops is
()
2
2
0
2
11
(
3
.
5
0
m
/
s
)
1.18 m
22
9.8 m/s
sin 32.0
v
x
a
§·
−
¨¸
= −
°
©¹
,
or   1.18 m.
x
=
(b) The time is
00
2
3.50m/s
0.674s.
sin
9.8m/s sin 32.0
t
ag
θ
−
==
−
=
−
=
°
(c) That the returnspeed is identical to the initial speed is to be expected since there are
no dissipative forces in this problem. In order to prove this, one approach is to set
x
= 0
and solve
xv
t a
t
0
1
2
2
for the total time (up and back down)
t
. The result is
t
v
a
v
g
−
°
=
23
5
0
98
320
135
sin
.
.s
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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, Force, Gravity

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