Chapter three Motion in two dimension
Example(2)
A particle with velocity r v o
= -
2 iˆ
+
4
ˆ
j at t=0 undergoes a constant acceleration r a of magnitude a=3 m/sec at angle q =
130 positive direction of the x-axis, what is the particle's velocity r v f from the
at t=5sec, in unit vector notation and in magnitude angle notation?
Solution: v x v y
= v o
= v o y x
+
+ a x t a y t a a y x
=
= a cos q a sin q
=
1 cos 130
=
3 sin 130
=
1 .
93 m
=
2 .
30 m
/ sec 2
/ sec 2
At t=5sec v x
= -
2
+
(
-
1 .
93 )( 5 )
= -
11 .
65 v y
=
4
+
2 .
30 ( 5 )
=
15 .
5 r v
= v x iˆ
+ v magnitude y jˆ of
= r v
-
11 .
65 iˆ m
+
/ sec
15 .
5 jˆ m
=
/ sec v
= v
= v 2 x
+ v 2 y
=
19 m / sec q = tan
-
1 ( v y v x
)
=
127 0
١

Chapter three Motion in two dimension
Example(3)
Ball kicked horizontally at 8m/sec off a 50m high cliff, find: (1)time to impact (2)speed at impact (3)impact point (4)angle at impact ?
y v
0
=
18 m / sec
H=50m
x b
Solution : x ( t )
= v
0 cos q t , q =
0
= v
0 t ..........
......( 1 ) y ( t )
= y
0
+ v
0 y t
-
1
2 gt 2
=
H
-
1
2 gt 2
At impact we
..........
must
.....( have y
2 )
=
0 .
, v
0 y
=
0 , y
0
=
H
Solving Eq .( 2 ) t
=
T
=
2 H
=
2 * 50 g 9 .
81 x ( T )
=
18 * 3 .
19
=
57 .
42
=
3 .
19 sec
V x
( T ) dx dt
= v
0
=
18 m / sec ( indpendent of time
V y
( T ) dy dt tan b =
V y
V x
=
=
gt
-
=
60
-
.
1 0
9 .
81 * 3 .
19
= -
31 .
26 m / sec v
= v 2 x
+ v 2 y
=
36 .
1 m / sec ( speed )
)
٢

Chapter three Motion in two dimension
Example(4)
Gun fired a bullet with velocity 200m/sec by an 40
0
with horizontal, find a velocity and position of a bullet after 20sec and find range and time required to return to ground?
Solution : v
0 x
= v
0 cos q =
200 cos 40
=
153 .
2 m / sec v
0 y v
0 x
= v
0
= v x sin
= q =
153 .
200 sin
2 m / sec
40 v y
= v
0 x
gt
=
128
At v y
= t
=
-
20
67 .
sec
4 m / sec
.
6
-
9 .
8 t
=
128 .
6 m / sec v
= v 2 x
+ v 2 y
=
167 m / sec t
=
2 v g
0 y
=
2 ( 128 .
6 )
9 .
8
R h
=
= v v 2
0 sin 2 q
=
9 .
8
2
0 sin
2 g
2 q
=
( 200 ) 2 sin 2 ( 40 )
( 200 ) 2 (sin 40 ) 2
2 *
9 .
8
9 .
8
=
4021 m
=
843 .
7 m
٣

Chapter three Motion in two dimension
Example(5)
A ball tied to the end of a sting 0.50m in length swings in a vertical circle under the influence of gravity as show in figure down, when the string makes an angle q =
20 0 with the vertical, the ball has a speed of
1.5m/sec. (1) find the magnitude of the radial component of acceleration at this instant (2)what is the magnitude of the tangential acceleration
(3) find the magnitude and direction of the total acceleration
Solution : q
v =0 r
(1) a = r v r
2
=
( 1 .
5 m / sec)
0 .
50 m
2
=
4 .
5 m / sec 2
g
a r
(2) a
= g sin q =
9 .
8 sin 20 0
=
3 .
4 m / sec 2 f
a
a t
(3) a = a 2 r
+ a t
2
=
( 4 .
5 ) 2
+
( 3 .
4 ) 2 m / sec 2
=
5 .
6 m / sec 2
If the angle between a and the string, then f = tan
-
1 a a t r
= tan
-
1
é
êë
3 .
4 m
4 .
5 m
/ sec 2
/ sec 2
ù
úû
=
37 0
٤