Projectile motion ( physics) surabay nsh

Projectile motion, physics

1. PROJECTILE MOTION
Senior High School Physics
Modified by:
Gladys D. Silvestre
Part 1 Part 2

2. Introduction
Projectile Motion:
Motion through the air without a propulsion
Examples:

3. Part 1.
Motion of Objects Projected
Horizontally

4. v
i
x
y

5. x
y

6. x
y

7. x
y

8. x
y

9. x
y
•Motion is accelerated
•Acceleration is constant,
and downward
• a = g = -9.81m/s
2

•The horizontal (x)
component of velocity is
constant
•The horizontal and vertical
motions are independent of
each other, but they have a
common time
g = -9.81m/s
2

10. ANALYSIS OF MOTION
ASSUMPTIONS:
• x-direction (horizontal): uniform motion
• y-direction (vertical): accelerated motion
• no air resistance
QUESTIONS:
• What is the trajectory?
• What is the total time of the motion?
• What is the horizontal range?
• What is the final velocity?

11. x
y
0
Frame of reference:
h
v
i
Equations of motion:
X
Uniform m.
Y
Accel. m.
ACCL. a
x = 0a
y = g = -9.81
m/s
2
VELC. v
x
= v
i
v
y
= g t
DSPL. x = v
i
ty = h + ½ g t
2
g

12. Trajectory
x = v
i
t
y = h + ½ g t
2
Eliminate time, t
t = x/v
i
y = h + ½ g (x/v
i
)
2
y = h + ½ (gx
2
/v
i
2
)
y = ½ (gx
2
/v
i
2
) + h
y
x
h
Parabola, open down
v
i1
v
i2 > v
i1

13. Total Time, Δt
y = h + ½ g t
2
final y = 0 y
x
h
t
i
=0
t
f =Δt
0 = h + ½ g (Δt)
2
Solve for Δt:
Δt = √ 2h/(-g)
Δt = √ 2h/(9.81ms
-2
)
Total time of motion depends
only on the initial height, h
Δt = t
f - t
i

14. Horizontal Range, Δx
final y = 0, time is
the total time Δt
y
x
h
Δt = √ 2h/(-g)
Δx = v
i √ 2h/(-g)
Horizontal range depends on the
initial height, h, and the initial
velocity, v
i
Δx
x = v
i t
Δx = v
i Δt

15. VELOCITY
v
v
x
= v
i
v
y
= g t
v = √v
x
2
+ v
y
2
v= √v
i
2
+g
2
t
2

16. FINAL VELOCITY
v
v
x
= v
0
v
y
= g t
v = √v
x
2
+ v
y
2
v = √v
i
2
+g
2
(2h /(-g))
v = √ v
i
2
+ 2h(-g)
Δt = √ 2h/(-g)

17. HORIZONTAL THROW - Summary
h – initial height, v
0 – initial horizontal velocity, g = -9.81m/s
2

18. Part 2.
Motion of objects projected at an
angle

19. v
i
x
y
θ
v
ix
v
iy
Initial position: x = 0, y = 0

20. x
y
•Motion is accelerated
•Acceleration is constant, and
downward
• a = g = -9.81m/s
2

•The horizontal (x) component of
velocity is constant
•The horizontal and vertical
motions are independent of each
other, but they have a common
time
a = g =
- 9.81m/s
2

21. ANALYSIS OF MOTION:
ASSUMPTIONS
• x-direction (horizontal): uniform motion
• y-direction (vertical): accelerated motion
• no air resistance
QUESTIONS
• What is the trajectory?
• What is the total time of the motion?
• What is the horizontal range?
• What is the maximum height?
• What is the final velocity?

22. Equations of motion:

23. Equations of motion:

24. Trajectory
Eliminate time, t
y
x
Parabola, open down
2
22
22
2
cos2
tan
cos2cos
sin
x
v
g
xy
v
gx
v
xv
y
i
ii
i







y = ax + bx
2

25. Total Time, Δt
final height y = 0, after time interval Δt
Solve for Δt:
(-g)
t = 0 Δt
x

26. Horizontal Range, Δx
final y = 0, time is
the total time Δt
x
Δx
y
0
Δt =
(-g)
Δx =
(-g)
Δx =
(-g)

27. Horizontal Range, Δx
Δx =
(-g)

28. Trajectory and horizontal range
2
22
cos2
tan x
v
g
xy
i 

0
5
10
15
20
25
30
35
0 20 40 60 80
15 deg
30 deg
45 deg
60 deg
75 deg
v
i
= 25 m/s

29. Velocity
•Final speed = initial speed (conservation of energy)
•Impact angle = - launch angle (symmetry of parabola)

30. Maximum Height
At maximum height v
y = 0
t
up
=
(-g)
t
up
= Δt/2
2(-g)
Recall:
Δt =
(-g)

31. Projectile Motion – Final Equations
Trajectory Parabola, open down
Total time Δt =
Horizontal range
Δx =
Max height
h
max =
(-g)
(-g)
2(-g)

32. PROJECTILE MOTION - SUMMARY
Projectile motion is motion with a constant
horizontal velocity combined with a constant
vertical acceleration
The projectile moves along a parabola

33. ACTIVITY:
An object is thrown horizontally from the top of the
cliff with an initial horizontal velocity of 25m/s. The
height of the cliff is 45 meter. Calculate the following:
a.) Time of Flight
b.) Horizontal Range
c.) Maximum Height

34. ACTIVITY:
A projectile is launched with an initial velocity of
40 m/s at an angle of 30 degrees above the
horizontal. Calculate the following:
a.) Time of Flight
b.) Horizontal Range
c.) Maximum Height