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# Chapter 6B – Projectile Motion

Chapter 6B – Projectile Motion. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University. © 2007. Describe the motion of a projectile by treating horizontal and vertical components of its position and velocity.

## Chapter 6B – Projectile Motion

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1. Chapter 6B – Projectile Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

2. Describe the motion of a projectile by treating horizontal and vertical components of its position and velocity. Objectives: After completing this module, you should be able to: • Solve for position, velocity, or time when given initial velocity and launch angle.

3. a = g W W W Projectile Motion A projectile is a particle moving near the Earth’s surface under the influence of its weight only (directed downward).

4. Simultaneously dropping yellowball and projecting red ball horizontally. Vertical and Horizontal Motion Click right to observe motion of each ball.

5. W W Vertical and Horizontal Motion Simultaneously dropping a yellowball and projecting a red ball horizontally. Why do they strike the ground at the same time? Once motion has begun, the downward weight is the only force on each ball.

6. Vertical Motion is the Same for Each Ball vox 0 s vx 1 s vy vx vy 2 s vy vy vx 3 s vy vy Ball Projected Horizontally and Another Dropped at Same Time:

7. Vertical Motion is the Same for Each Ball vox 0 s 1 s 2 s 3 s Observe Motion of Each Ball

8. Compare Displacements and Velocities vox 1 s 2 s 3 s 0 s 0 s 1 s vy vx 2 s vy vx 3 s vy Consider Horizontal and Vertical Motion Separately: vx Horizontal velocity doesn’t change. Vertical velocity just like free fall.

9. For the special case of horizontal projection: Horizontal displacement: Vertical displacement: Displacement Calculations for Horizontal Projection: For any constant acceleration:

10. For the special case of a projectile: Horizontal velocity: Vertical velocity: Velocity Calculations for Horizontal Projection (cont.): For any constant acceleration:

11. x +50 m 25 m/s -19.6 m y Example 1:A baseball is hit with a horizontal speed of 25 m/s. What is its position and velocity after 2 s? First find horizontal and vertical displacements: x = 50.0 m y = -19.6 m

12. 25 m/s vx vy Example 1 Cont.):What are the velocity components after 2 s? v0x = 25 m/s v0y = 0 Find horizontal and vertical velocity after 2 s: vx= 25.0m/s vy= -19.6 m/s

13. vo voy q vox Consider Projectile at an Angle: A red ball is projected at an angle q. At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction). vx = vox = constant Note vertical and horizontal motions of balls

14. Thus, the displacement components x and y for projectiles are: Displacement Calculations For General Projection: The components of displacement at time t are: For projectiles:

15. Thus, the velocity components vx and vy for projectiles are: Velocity Calculations For General Projection: The components of velocity at time t are: For projectiles:

16. vo voy q vox Displacement: Velocity: Problem-Solving Strategy: • Resolve initial velocity vo into components: 2. Find components of final position and velocity:

17. R y q x vo voy q vox Problem Strategy (Cont.): 3. The final position and velocity can be found from the components. 4. Use correct signs - remember g is negative or positive depending on your initial choice.

18. voy 160 ft/s Example 2:A ball has an initial velocity of 160 ft/s at an angle of 30o with horizontal. Find its position and velocity after 2 s and after 4 s. 30o vox Since vx is constant, the horizontal displacements after 2 and 4 seconds are: x = 277 ft x = 554 ft

19. voy 160 ft/s 2 s 4 s 30o vox 277 ft 554 ft x2 = 277 ft x4 = 554 ft Example 2: (Continued) Note: We know ONLY the horizontal location after 2 and 4 s. We don’t know whether it is on its way up or on its way down.

20. g = -32ft/s2 voy= 80ft/s 160 ft/s y2 y4 q 0 s 1 s 2 s 3 s 4 s Example 2 (Cont.):Next we find the vertical components of position after 2 s and after 4 s. The vertical displacement as function of time: Observe consistent units.

21. g = -32ft/s2 voy= 80 ft/s 160 ft/s y2 y4 q 0 s 1 s 2 s 3 s 4 s (Cont.)Signs of y will indicate location of displacement (above + or below – origin). 96 ft 16 ft Vertical position: Each above origin (+)

22. voy 160 ft/s 30o vox (Cont.):Next we find horizontal and vertical components of velocity after 2 and 4 s. Since vx is constant, vx = 139ft/s at all times. Vertical velocity is same as if vertically projected: At any time t:

23. Example 2: (Continued) g = -32ft/s2 vy= 80.0ft/s v2 160 ft/s v4 q 0 s 1 s 2 s 3 s 4 s At any time t: v2y = 16.0 ft/s v4y = -48.0 ft/s

24. g = -32ft/s2 vy= 80.0ft/s v2 160 ft/s v4 q 0 s 1 s 2 s 3 s 4 s Example 2: (Continued) Moving Up +16ft/s Moving down -48 ft/s The signs ofvy indicate whether motion is up (+) or down (-) at any time t. At 2 s: v2x = 139 ft/s; v2y = + 16.0 ft/s At4 s: v4x = 139 ft/s; v4y = - 48.0 ft/s

25. y2 = 96ft x2= 277 ft (Cont.):The displacement R2,q is found from the x2 and y2component displacements. t = 2 s R2 q 0 s 2 s 4 s R2= 293 ft q2= 19.10

26. t = 4 s R4 y4 = 64ft q 0 s x4= 554 ft 4 s (Cont.):Similarly, displacement R4,q is found from the x4 and y4component displacements. R4= 558 ft q4= 6.590

27. v2 g= -32ft/s2 voy= 80.0ft/s 160 ft/s Moving Up +16 ft/s q 0 s 2 s (Cont.): Now we find the velocity after 2 s from the components vx and vy. v2x = 139 ft/s v2y = + 16.0 ft/s v2 = 140 ft/s q2= 6.560

28. g= -32ft/s2 voy= 80.0ft/s 160 ft/s v4x = 139 ft/s v4y = - 48.0 ft/s v4 q 0 s 4 s (Cont.) Next, we find the velocity after 4 s from the components v4x and v4y. v4= 146 ft/s q2= 341.70

29. voy 28 m/s vy= 0 ymax 30o vox Example 3: What are maximum height and range of a projectile if vo = 28 m/s at 300? vox = 24.2 m/s voy = + 14 m/s Maximum y-coordinate occurs when vy= 0: ymaxoccurs when14 – 9.8t = 0or t = 1.43 s

30. voy vox = 24.2 m/s 28 m/s vy= 0 ymax voy = + 14 m/s 30o vox Example 3(Cont.): What is maximum height of the projectile if v = 28 m/s at 300? Maximum y-coordinate occurs whent = 1.43 s: ymax= 10.0m

31. voy vox = 24.2 m/s 28 m/s voy = + 14 m/s vox 30o Range xr Example 3(Cont.): Next, we find the range of the projectile if v = 28 m/s at 300. The range xr is defined as horizontal distancecoinciding with the time for vertical return. The time of flight is found by setting y = 0: (continued)

32. voy vox = 24.2 m/s 28 m/s voy = + 14 m/s vox 30o Range xr Example 3(Cont.): First we find the time of flight tr, then the range xr. (Divide by t) xr= 69.2m xr = voxt = (24.2 m/s)(2.86 s);

33. R 0 1.2 m 2 m Example 4: A ball rolls off the top of a table 1.2 m high and lands on the floor at a horizontal distance of 2 m. What was the velocity as it left the table? Note: x = voxt = 2 m y = voyt + ½ayt2 = -1.2 m First find t from y equation: ½(-9.8)t2= -(1.2) t = 0.495 s

34. Note: x = voxt = 2 m R y = ½gt2 = -1.2 m 1.2 m 2 m Example 4 (Cont.): We now use horizontal equation to find voxleaving the table top. Use t = 0.495 s in x equation: The ball leaves the table with a speed: v = 4.04 m/s

35. 0 1.2 m vx 2 m vy Example 4 (Cont.): What will be its speed when it strikes the floor? Note: t = 0.495s vx = vox = 4.04m/s vy = vy + gt vy= -4.85 m/s vy = 0 + (-9.8 m/s2)(0.495 s) v4= 146 ft/s q2= 309.80

36. y = 0; a = -9.8 m/s2 Initial vo: vo =25 m/s vox = vo cos q Time of flight t 600 voy = vo sin q Example 5.Find the “hang time” for the football whose initial velocity is 25 m/s, 600. Vox = (25 m/s) cos 600; vox = 12.5 m/s Voy = (25 m/s) sin 600; vox = 21.7 m/s Only vertical parameters affect hang time.

37. y = 0; a = -9.8 m/s2 Initial vo: vo =25 m/s vox = vo cos q Time of flight t 600 voy = vo sin q Example 5 (Cont.)Find the “hang time” for the football whose initial velocity is 25 m/s, 600. 4.9 t2 = 21.7 t 4.9 t = 21.7 t = 4.42 s

38. v = 11 m/s q =300 Example 6. A running dog leaps with initial velocity of 11 m/s at 300. What is the range? Draw figure and find components: voy = 11 sin 300 vox = 9.53 m/s voy = 5.50 m/s vox = 11 cos 300 To find range, first find t when y = 0; a = -9.8 m/s2 4.9 t2 = 5.50 t t = 1.12 s 4.9 t = 5.50

39. v = 10 m/s q =310 Example 6 (Cont.) A dog leaps with initial velocity of 11 m/s at 300. What is the range? Range is found from x-component: voy = 10 sin 310 vx = vox = 9.53 m/s x = vxt; t = 1.12s vox = 10 cos 310 Horizontal velocity is constant: vx = 9.53m/s x = (9.53 m/s)(1.12 s) = 10.7 m Range: x = 10.7 m

40. Summary for Projectiles: 1. Determine x and y components v0 2. The horizontal and vertical components of displacement at any time t are given by:

41. Summary (Continued): 3. The horizontal and vertical components of velocity at any time t are given by: 4. Vector displacement or velocity can then be found from the components if desired:

42. CONCLUSION: Chapter 6B Projectile Motion

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