1. Introduction to 2D Projectile Motion

2. Projectile Motion An example of 2-dimensional motion.
Something is fired, thrown, shot, or hurled near the earth’s surface.
Horizontal velocity is constant.
Vertical velocity is accelerated.
Air resistance is ignored.

3. Trajectory of Projectilex y
This projectile is launched at an angle and rises to a peak before falling back down.

4. Trajectory of Projectilex y
The trajectory of such a projectile is defined by a parabola.

5. Trajectory of Projectilex y
Range
The RANGE of the projectile is how far it travels horizontally.

6. Trajectory of Projectilex y
Maximum
Height
Range
The MAXIMUM HEIGHT of the projectile occurs halfway through its range.

7. Trajectory of Projectilex y g g g g g
Acceleration points down at 9.8 m/s2 for the entire trajectory.

8. Position graphs for 2-D projectilesx y t

9. To work projectile problems…
…you must first resolve the initial velocity into components.
Vo,y = Vo sin  Vo 
Vo,x = Vo cos 

10. Trajectory of Projectilex y v v v vo vf
Velocity is tangent to the path for the entire trajectory.

11. Trajectory of Projectilex y vx vy vx vy vx vy vx vx vy
The velocity can be resolved into components all along its path.

12. Trajectory of Projectilex y vx vy vx vy vx vy vx vx vy
Notice how the vertical velocity changes while the horizontal velocity remains constant.

13. Trajectory of Projectilex y vx vy vx vy vx vy vx vx vy
Where is there no vertical velocity?

14. Trajectory of Projectilex y vx vy vx vy vx vy vx vx vy
Where is the total velocity maximum?

15. 2D Motion Resolve vector into components.
Position, velocity or acceleration
Work as two one-dimensional problems.
Each dimension can obey different equations of motion.

16. Horizontal Component of Velocity
Newton's 1st Law
Is constant
Not accelerated
Not influence by gravity
Follows equation:
x = Vo,xt

17. Horizontal Component of Velocity

18. Vertical Component of Velocity
Newton's 2nd Law
Undergoes accelerated motion
Accelerated by gravity (9.8 m/s2 down)
Vy = Vo,y - gt
y = yo + Vo,yt - 1/2gt2
Vy2 = Vo,y2 - 2g(y – yo)

19. Horizontal and Vertical

20. Horizontal and Vertical

21. Launch angle vo
Zero launch angle

22. Launch angle  vo
Positive launch angle

23. Symmetry in Projectile Motion vo - vo
Negligible air resistance
Projectile fired over level ground
Launch and Landing Velocity

24. Symmetry in Projectile Motion
to = 0
Time of flight

25. Symmetry in Projectile Motion
to = 0 2t
Projectile fired over level ground
Time of flight
Negligible air resistance

26. Problem Solution Strategy:
1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = m/s2
2. Resolve the initial velocity vo into its x and y components:
vox = vo cos θ voy = vo sin θ

27. Problem Solution Strategy (pt 2)
3. The horizontal and vertical components of its position at any instant is given by: x = voxt y = voy t +½gt2
4. The horizontal and vertical components of its velocity at any instant are given by: vx = vox vy = voy + gt

28. Problem Solution Strategy (pt 3)
5. The final position and velocity can then be obtained from their components.