ilide.info-motion-in-two-dimension-pr_d38aadd2be4949ed06275d8691959e74.pdf

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3. -objects move both in vertical (y
coordinate) and horizontal (x
coordinate) directions
at the same time.
X
Y
P Y S I C SH

4. -is the motion of the object thrown or
projected into the air, and it moves along a
curve path or parabolaunder the influence of
gravity only. The object is called the projectile,
and its path is called thetrajectory.
NOTE: we will assume that the air resistance is
negligible.
S I C SP H Y

5. since the gravitational force pulls downward,
gravity will only affect the vertical component of the
velocity (Vy). The horizontal component of the
velocity (Vx) will remain unaffected and stays
constant.
X
YX
Y
S I C SP H Y

6. -analyze the motion in each direction
separately. We will use one set of equations
to describe the horizontal motion and
another set of equations to describe the
vertical motion.
S I C SP H Y

7. S I C SP H Y
The Horizontal and Vertical Components of Projectile Motion
Quantity Horizontal Vertical
Acceleration a
x= 0 a
y= g
Velocity
v
0x= v
0cosѲ v
0y= v
0sinѲ
v
x= v
0x v
y= v
0y+ a
yt
Displacement d
x= v
xt
d
y= v
0yt +
1
/
2a
yt
2
2a
yd
y= v
y
2
–v
0y
2

8. x = horizontal distance (m)
y= vertical distance (m)
v= velocity (m/s)
Vx= horizontal velocity (m/s)
Vy= vertical velocity (m/s)
Vox= initial horizontal velocity (m/s)
Voy= initial vertical velocity (m/s)
t= time (s)
g/ay= acceleration to gravity (9.8 m/s²)
S I C SP H Y

9. Velocity is a vector quantity and must be broken into
two components with the use of trigonometric
functions.
Sine =
opposite
hypotenuse
Cosine =
adjacent
hypotenuse
Tangent =
opposite
adjacent
S I C SP H Y

10. V = Voy² + Vox²
S I C SP H Y

11. Problem1:A body is projected with a velocity of 20 m/sat
50
o
to the horizontal. Find
a. Maximum height reached
b. Time of flight
c. Range of the projectile and
d. Magnitude of the velocity before it hits the ground
20m/s
50°
Hmax= ?
r = ?
Time of flight = ?
Magnitude of velocity before
it hits the ground = ?
S I C SP H Y

12. Problem1:A body is projected with a velocity of 20 m/sat
50
o
to the horizontal.
HorizontalComponents VerticalComponents
Vx= cosØVo
Vx=cos50°(20m/s)
Vox= 12.86 m/s
Voy= sinØVo
Voy= sin50°(20m/s)
Voy= 15.32m/s
g= 9.8 m/s²
Voy= 15.32 m/s
Vox= 18.13 m/s
S I C SP H Y

13. Problem 1: An object is launched at a velocity of 20
m/s in a direction making an angle of 50°upward
with the horizontal.
a. Maximum height reached
d = Vy
2
-Voy²
2g
= 0-(15.32m/s)²
2(-9.8 m/s²)
= -234.70 m²/s²
-19.6 m/s²
maximum height=
11.97 m
S I C SP H Y

14. Problem 1: An object is launched at a velocity of 20
m/s in a direction making an angle of 50°upward
with the horizontal.
b. Time of flight
t = Vy-Voy
g
= 0-15.32 m/s
-9.8 m/s²
t = 1.56 s
Tt= 2 (1.56 s)
Total time= 3.126 s
S I C SP H Y

15. Problem 1: An object is launched at a velocity of 20
m/s in a direction making an angle of 50°upward
with the horizontal.
c. Range of the projectile and
R = VoxTt
= (12.86 m/s) (3.126 s)
Range = 40.2 m
S I C SP H Y

16. Problem 1: An object is launched at a velocity of 20 m/s in a
direction making an angle of 50°upward with the horizontal.
d. Magnitude of the velocity before it hits the
ground
Vg = Voy² + Vox²
= (15.32 m/s)² + (12.86 m/s)²
= 234.70 m²/s² + 165.37 m²/s²
= 400 m²/s²
Vg = 20 m/s
S I C SP H Y

17. Problem 1: An object is launched at a velocity of 20 m/s in a
direction making an angle of 50°upward with the horizontal.
Hmax= 11.97 m
Range = 40.2 m
t = 1.56 s
Total time=3.126 s
Vg = 20 m/s
Vo =20m/s
50
S I C SP H Y

18. Problem 2: A canon fires horizontally with a speed of 8.31 m/s
from the top of the cliff of height 23.0 m. Find
a. Time of flight
b. Range of the projectile
Vx= 8.31 m/s
H = -23.0 m
R = ?
S I C SP H Y

19. Problem 2: A canon fires horizontally with a speed of 8.31 m/s
from the top of the cliff of height 23.0 m. Find
a. Time of flight
b. Range of the projectile
HorizontalComponents VerticalComponents
Vx= 8.31 m/s Vy= 0
g= -9.8 m/s²
H= -23.0 m
S I C SP H Y

20. Problem 2: A canon fires horizontally with a speed of 8.31 m/s
from the top of the cliff of height 23.0 m. Find
a. Time of flight
b. Range of the projectile
T = 2d
g
= 2(-23.0m)
-9.8 m/s²
= -46 m
-9.8 m/s²
= 4.69 s²
T= 2.17 s
S I C SP H Y

21. Problem 2: A canon fires horizontally with a speed of 8.31 m/s
from the top of the cliff of height 23.0 m. Find
a. Time of flight
b. Range of the projectile
R = VxT
= (8.31 m/s) (2.17 s)
Range = 18.0327 m
S I C SP H Y

22. Problem 2: A canon fires horizontally with a speed of 8.31 m/s
from the top of the cliff of height 23.0 m. Find
a. Time of flight
b. Range of the projectile
H = -23.0 m
Vx= 8.31 m/s
Time of flight = T= 2.17 s
Range = 18.0327 m
S I C SP H Y

23. Exercises:
1.A football is kicked with a velocity of 20.0 m/s
at an angle of 30
0
with the horizontal. Find:
a.The horizontal and vertical components of its
initial velocity
b.The time it took to reach its maximum height
c.The time of flight
d.The maximum height it reached and
e.The range it traveled
S I C SP H Y

24. Exercises:
1.A football is kicked with a velocity of 20.0 m/s at an
angle of 30
0
with the horizontal. Find:
a.The horizontal and vertical components of its
initial velocity
S I C SP H Y
Vox= cosѲVo
= cos30(20m/s)
= 17.32 m/s
Voy= sinѲVo
= sin 30(20m/s)
= 10 m/s

25. Exercises:
1.A football is kicked with a velocity of 20.0 m/s at an
angle of 30
0
with the horizontal. Find:
b. The time it took to reach its maximum height
c. The time of flight
S I C SP H Y
t =
??????�−????????????�
??????
=
0−(10
??????
??????
)
−9.8??????/�
2
= 1.02 s
T = 2t
= 2 (1.02s)
= 2.04s

26. Exercises:
1.A football is kicked with a velocity of 20.0 m/s
at an angle of 30
0
with the horizontal. Find:
d. The maximum height it reached and
S I C SP H Y
??????=
??????�
2
−????????????�
2
2??????
=
0−10??????/�
2
2(−9.8
??????
??????
2
)
= -100 m
2
/s
2
-19.6 m/s
2
dy= 5.10 m

27. Exercises:
1.A football is kicked with a velocity of 20.0 m/s
at an angle of 30
0
with the horizontal. Find:
e. The range it traveled
S I C SP H Y
R = Vxt
= (17.32 m/s)(2.04s)
= 35.33 m

28. Exercises:
2.A ball thrown horizontally from the edge of the
top of a building 49.0 m high strikes the ground
24.5 m from the foot of the building. Find:
a.The time it takes the ball to reach the ground
b.The initial velocity of the ball, and
c.The velocity just before the ball strikes the
ground.
S I C SP H Y

29. Exercises:
2.A ball thrown horizontally from the edge of the top of a
building 49.0 m high strikes the ground 24.5 m from
the foot of the building. Find:
a.The time it takes the ball to reach the ground
S I C SP H Y
T =
2??????
??????
=
2(−49.0??????)
−9.8??????/�
2
=
−98??????
−9.8??????/�
2
= 3.16 s

30. Exercises:
2.A ball thrown horizontally from the edge of the top of a
building 49.0 m high strikes the ground 24.5 m from
the foot of the building. Find:
b. The initial velocity of the ball, and
S I C SP H Y
Vx =
??????�
�
=
24.5??????
3.16�
= 7.75 m/s

31. Exercises:
2.A ball thrown horizontally from the edge of the top of a
building 49.0 m high strikes the ground 24.5 m from the foot
of the building. Find:
c. The velocity just before the ball strikes the ground.
S I C SP H Y
Vy= Voy+ gt
= 0 + (-9.8 m/s
2
)(3.16 s)
= -30.97 m/s
??????=??????�
2
+??????�
2
= 7.75??????/??????
2
+−30.97??????/??????
2
= 31.92 m/s

32. Exercises:
3.John kicks the ball and ball does projectile
motion with an angle of 53
0
to horizontal. Its
initial velocity is 10m/s. Find:
a.The maximum height it can reach
b.The horizontal displacement
c.The total time required for this motion
S I C SP H Y

33. THANK YOU !!!
S I C SP H Y

34. DESCRIPTION FORMULA
HorizontalDistance X = Vxt
X=Vo²sin2Ø
g
Horizontal velocity Vx= Vox
Vertical distance y= Voyt-1/2gt²
Vertical velocity c
Time of flight t= 2VosinØ
2g
t = Voy t = 2H
g g
Maximum heightreached Hmax= Vo²sinØ
g
Hmax= Voy²
2g
S I C SP H Y