1. PHYS 20 LESSONS Unit 2: 2-D Kinematics Projectiles Lesson 5: 2-D Projectiles

2. Reading Segment #1: Horizontal Projectiles To prepare for this section, please read: Unit 2: p.18

3. E. 2-D Projectiles When a projectile is thrown, the path of the object is a parabola. i.e.

4. Animation: Pirate Ship http://library.thinkquest.org/27585/lab/sim_pirates.html

5. To study 2-D projectile motion, we will analyze two situations: 1. Projectiles launched horizontally 2. Projectiles launched at an angle We will assume there is no air resistance for all examples.

6. E1. Projectiles launched horizontally Consider a cannonball fired horizontally off of a cliff. We will analyze this motion from the horizontal (x) and the vertical (y) perspectives separately.

7. Horizontal (x) perspective: Because there is no air resistance, there is no horizontal acceleration. It will move with a constant horizontal velocity (much like a cannonball that experiences no gravity at all).

8. Vertical (y) perspective: It has no vertical velocity at the start, but it is accelerating downward at 9.81 m/s 2. So, it will speed up vertically (from rest).

9. Animation: Horizontally-Launched Projectile http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/hl p.html

10. Summary: Horizontal Motion Moves at a constant velocity Use the formula v = d t Vertical Motion Starts from rest (v i = 0). Accelerates at 9.81 m/s 2 downward. Use the kinematic equations.

11. Ex. 1An object is launched horizontally off of a cliff that is 18.0 m high. How far does it travel horizontally before it hits the ground. Assume no air resistance and level ground.

12. 14.0 m/s 18.0 m d ? We will analyze the vertical motion first. From this perspective, it falls (from rest) a height of 18.0 m.

13. Vertical (y):Ref: Down + Up - v i = 0 v f a = 9.81 m/s 2 d = 18.0 m t ? We need to find the time of the flight. Use the equation d = v i t + 0.5 a t 2

14. d = v i t + 0.5 a t 2 d = t 2 0.5 a t = d = 18.0 m 0.5 a 0.5 (9.81 m/s 2 ) = 1.916 s We can now use this flight time for the horizontal motion.

15. Horizontal (x) Motion: It moves at a constant horiztonal speed of 14.0 m/s. So,v = d t d = v t = (14.0 m/s) (1.916 s) = 26.8 m

16. Practice Problems Try these problems in the Physics 20 Workbook: Unit 2 p. 19 #1 - 7

17. Reading Segment #2: Diagonal Projectiles To prepare for this section, please read: Unit 2: p.7

18. E2. Projectiles launched diagonally Consider a cannonball fired diagonally. v We will analyze this motion from the horizontal (x) and the vertical (y) perspectives separately.

19. Horizontal (x) perspective: v v x The horizontal component of the velocity (v x ) will remain constant throughout the motion. There is no horizontal acceleration.

20. Vertical (y) perspective: v Vertically, the object will accelerate downward at 9.81 m/s 2. Thus, its speed will decrease on the way up, and its speed will increase on the way down.

21. Vertical (y) perspective: v Notice, at the top of the motion (max height), although the vertical component is zero, it is not at rest. There is still a horizontal component.

22. Animation: Diagonally-launched projectile: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/nh lp.html

23. Summary (Diagonal Projectiles) Horizontally: Moves at a constant velocity. Use v x = d / t Vertically: Accelerates downward at 9.81 m/s 2 Recall (symmetry) At the same height, v up = v down If it returns to the same height, t up = t down

24. Ex. 2A golf ball is hit with a speed of 78.0 m/s at 13.0  above the horizontal. Find: a) the maximum horizontal distance the ball travels while it is in the air b) the maximum height of the ball Assume no air resistance and level ground.

25. Step 1: Find the x- and y-components of the initial velocity y 78.0 m/s v y 13.0  v x x = 76.00 m/s= 17.546 m/s

26. a) Vertical (y) perspective: From the vertical perspective, the ball starts with an upward velocity of 17.546 m/s. 17.546 m/s It goes up and then comes back down to the same height. 17.546 m/s It will have the same speed when it returns to the same height.

27. a) Vertical (y) perspective:Ref: Up + ListDown - v i = +17.546 m/s v f = -17.546 m/s a = -9.81 m/s 2 d = t = ? 17.546 m/s We need to find the total time the ball is in the air. 17.546 m/s

28. a) Vertical (y) perspective:Ref: Up + ListDown - v i = +17.546 m/s v f = -17.546 m/s a = -9.81 m/s 2 d = t = ? Equation: a = v f - v i t

29. a = v f - v i t a t = v f - v i t = v f - v i a = (-17.546 m/s) - (+17.546 m/s) -9.81 m/s 2 = 3.5772 s

30. Horizontal (x) Perspective: The ball travels at a constant speed of 76.00 m/s (v x ) for a total time of 3.5772 s. v x = d t d = v x t = (76.00 m/s) (3.5772 s) = 272 m

31. b) Vertical (y) perspective: When the ball reaches its maximum height, its vertical velocity is zero (even though it is still moving horizontally). Rest 17.546 m/s

32. b) Vertical (y) perspective:Ref: Up + ListDown - v i = +17.546 m/s v f = 0 Rest a = -9.81 m/s 2 d = t = ? 17.546 m/s Equation: v f 2 = v i 2 + 2 a d

33. v f 2 = v i 2 + 2 a d v f 2 - v i 2 = 2 a d d = v f 2 - v i 2 2a = 0 - (17.546 m/s) 2 2 (-9.81 m/s 2 ) = 15.7 m

34. Additional Animations: Cannon (target): http://zebu.uoregon.edu/nsf/cannon.html Diagonal Projectiles:  http://www.hazelwood.k12.mo.us/~grichert/projectile/proj.htmlhttp://www.hazelwood.k12.mo.us/~grichert/projectile/proj.html  http://www.msu.edu/user/brechtjo/physics/cannon/cannon.htmlhttp://www.msu.edu/user/brechtjo/physics/cannon/cannon.html

35. Practice Problems Try these problems in the Physics 20 Workbook: Unit 2 p. 22 #1 - 5