1. Two dimensional motion: PROJECTILE MOTION
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. Projectile Motion
A red marble is dropped off a cliff at the same time a black one is shot horizontally. At any point in time the marbles are at the same height, i.e., they’re falling down at the same rate, and they hit the ground at the same time. Gravity doesn’t care that the black ball is moving sideways; it pulls it downward just the same. Since gravity can’t affect horiz. motion, the black particle continues at a constant rate. With every unit of time, the marbles’ vertical speed increases, but their horiz. speed remains the same (ignoring air resistance).
continued on next slide

5. Projectile Motion
9.8 m/s2
Gravity’s downward pull is independent of horizon motion. So, the vertical acceleration of each marble is -g (for the whole trip), and the sideways acceleration of each is zero. (Gravity can’t pull sideways). Whatever horizon velocity the black one had when shot is a constant throughout its trip. Only its vertical velocity changes. (A vertical force like gravity can only produce vertical acceleration.)
9.8 m/s2
9.8 m/s2
continued on next slide

6. Projectile Motion (cont.)
vy = 0
t = 1
vy = 1
If after one unit of time the marbles have one unit of speed downward, then after two units of time they have two units of speed downward, etc. This follows directly from vf = v0 + a t. Since v0 = 0, downward speed is proportional to time. Note: The vectors shown are vertical components of velocity. The shot marble has a horizontal component too (not shown); the dropped one doesn’t.
t = 2
vy = 2
t = 3
vy = 3
vy = 4
t = 4
continued on next slide

7. Projectile Motion (cont.)
vx = v
Projectile Motion (cont.)
t = 0
t = 1
t = 2
t = 3
t = 4 vx vy
Since the shot black marble experiences no horiz. forces (ignoring air), it undergoes no horiz. acceleration. Therefore, its horiz. velocity, doesn’t change. So, the horiz. vector has a constant magnitude, but the vertical vector gets longer. The resultant (the net velocity vector in blue) gets longer and points more downward with time. When t = 0, v = vx for the shot marble. v = vy for the dropped marble for the whole trip. v vx vy vy v vx vy vy v
continued on next slide vx

8. Projectile Motion (cont.)v
The trajectory of any projectile is parabolic. (We’ll prove this later.) If its initial velocity vector is horizontal, as with the black marble, the launch site is at the vertex of the parabola. v
The velocity vector at any point in time is tangent to the parabolic trajectory. Moreover, velocity vectors are always tangent to the trajectory of any moving object, regardless of its shape. v
continued on next slide

9. Projectile Motion (cont.)
y = 1
The vertical displacements over consecutive units of time show the familiar ratio of odd numbers that we’ve seen before with uniform acceleration. Measured from the starting point, the vertical displacements would be 1, 4, 9, 16, etc., (perfect squares), but the horiz. displacements form a linear sequence since there is no acceleration in that direction.
y = 3
x = 1
x = 2
y = 5
x = 3
y = 7
x = 4
continued on next slide

10. y v0 x

11. y x

12. y x

13. y x

14. a = -g = -9.8m/s2 y Motion is accelerated
Acceleration is constant, and downward
a = g = -9.81m/s2
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 = -9.8m/s2 x

15. 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?

16. X Uniform m. Y Accel. m. ACCL. ax = 0 ay =- g = -9.81 m/s2 VELC.
Frame of reference:
Equations of motion: x y X
Uniform m. Y
Accel. m.
ACCL.
ax = 0
ay =- g = m/s2
VELC.
vx = v0
vy = -g t
DSPL.
x = v0 t
y = h - ½ g t2 v0 h g

17. Trajectory y = h - ½ (g/v02) x2 y = -½ (g/v02) x2 + h x = v0 t
y = h - ½ g t2
Parabola, open down
Eliminate time, t
v02 > v01
v01
t = x/v0
y = h - ½ g (x/v0)2
y = h - ½ (g/v02) x2
y = -½ (g/v02) x2 + h

18. Total Time, Δt Δt = √ 2h/g Δt = √ 2h/(9.81ms-2) y = h - ½ g t2
Δt = tf - ti
y = h - ½ g t2 y x h
final y = 0
0 = h - ½ g (Δt)2
ti =0
Solve for Δt:
Δt = √ 2h/g
Δt = √ 2h/(9.81ms-2)
tf =Δt
Total time of motion depends only on the initial height, h

19. Horizontal Range, Δx Δx = v0 Δt Δt = √ 2h/g Δx = v0 √ 2h/g Δx x = v0 ty x h
final y = 0, time is the total time Δt
Δx = v0 Δt
Δt = √ 2h/g
Δx = v0 √ 2h/g Δx
Horizontal range depends on the initial height, h, and the initial velocity, v0

20. VELOCITY Θ v = √v02+g2t2 tan Θ = = vx = v0 vy = - g t v = √vx2 + vy2

21. FINAL VELOCITY Δt = √ 2h/g Θ v v = √v02+g2(2h /g) v = √ v02+ 2hg
vx = v0
Δt = √ 2h/g v
tan Θ = - g Δt / v0
= -(g)√2h/g / v0
= -√2hg / v0 Θ
vy = - g t
v = √vx2 + vy2
v = √v02+g2(2h /g)
v = √ v02+ 2hg
Θ is negative (below the horizontal line)

22. HORIZONTAL THROW - Summary
h – initial height, v0 – initial horizontal velocity, g = 9.81m/s2
Trajectory
Half -parabola, open down
Total time
Δt = √ 2h/g
Horizontal Range
Δx = v0 √ 2h/g
Final Velocity
v = √ v02+ 2hg
tan Θ = -√2hg / v0

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

24. x y vi
Initial position: x = 0, y = 0
vix
viy
Initial velocity: vi = vi [Θ]
Velocity components:
x- direction : vix = vi cos Θ
y- direction : viy = vi sin Θ θ

25. y a = - g = - 9.81m/s2 x Motion is accelerated
Acceleration is constant, and downward
a = g = -9.81m/s2
The horizontal (x) component of velocity is constant
The horizontal and vertical motions are independent of each other, but they have a common time x

26. 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?

27. X Uniform motion Y Accelerated motion
Equations of motion: X
Uniform motion Y
Accelerated motion
ACCELERATION
ax = 0
ay = - g = m/s2
VELOCITY
vx = vix= vi cos Θ
vx = vi cos Θ
vy = viy- g t
vy = vi sin Θ - g t
DISPLACEMENT
x = vix t = vi t cos Θ
x = vi t cos Θ
y = h + viy t - ½ g t2
y = vi t sin Θ - ½ g t2

28. X Uniform motion Y Accelerated motion
Equations of motion: X
Uniform motion Y
Accelerated motion
ACCELERATION
ax = 0
ay = - g = m/s2
VELOCITY
vx = vi cos Θ
vy = vi sin Θ - g t
DISPLACEMENT
x = vi t cos Θ
y = vi t sin Θ - ½ g t2

29. Trajectory x = vi t cos Θ y = vi t sin Θ - ½ g t2 Parabola, open down
Eliminate time, t
t = x/(vi cos Θ) 2
cos
tan
sin x v g y gx i Q - =
y = bx - ax2 x

30. Total Time, Δt y = vi t sin Θ - ½ g t2 Δt =
final height y = 0, after time interval Δt
0 = vi Δt sin Θ - ½ g (Δt)2
Solve for Δt: x
0 = vi sin Θ - ½ g Δt
Δt =
2 vi sin Θ
t = Δt g

31. Horizontal Range, Δx Δx = vi Δt cos Θ Δt = Δx = Δx = x = vi t cos Θ Δxy
final y = 0, time is the total time Δt
Δx = vi Δt cos Θ
Δt =
2 vi sin Θ g x
sin (2 Θ) = 2 sin Θ cos Θ Δx
Δx =
2vi 2 sin Θ cos Θ g
Δx =
vi 2 sin (2 Θ) g

32. Horizontal Range, Δx Δx = vi 2 sin (2 Θ) Θ (deg) sin (2 Θ) g
0.00 15
0.50 30
0.87 45
1.00 60 75 90
CONCLUSIONS:
Horizontal range is greatest for the throw angle of 450
Horizontal ranges are the same for angles Θ and (900 – Θ)

33. Trajectory and horizontal range2
cos
tan x v g y i Q - =
vi = 25 m/s

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

35. Maximum Height vy = vi sin Θ - g t y = vi t sin Θ - ½ g t2
At maximum height vy = 0
0 = vi sin Θ - g tup
hmax = vi t upsin Θ - ½ g tup2
hmax = vi2 sin2 Θ/g - ½ g(vi2 sin2 Θ)/g2
hmax =
vi2 sin2 Θ 2g
tup =
vi sin Θ g
tup = Δt/2

36. Projectile Motion – Final Equations
(0,0) – initial position, vi = vi [Θ]– initial velocity, g = 9.8 m/s2
Trajectory
Parabola, open down
Total time
Δt =
Horizontal range
Δx =
Max height
hmax =
2 vi sin Θ g
vi 2 sin (2 Θ) g
vi2 sin2 Θ
2 g

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