Objects launched are projectiles balls, bullets, arrows, space ships…

2. Objects launched are projectiles • balls, bullets, arrows, space ships… • The PATH a projectile follows is the TRAJECTORY • The trajectory for a projectile is a PARABOLA We will ignore wind resistance

3. Motion for a Projectile is described in terms of… • Position • Velocity • Acceleration • All three of these are vector quantities and will be represented with arrows like the previous topic • We must remember that horizontal and verticalvelocities (components) of a projectile are INDEPENDENT!

4. Drop a ball Accelerates Downward

5. Roll a Ball Rolls across Constant Velocity (No friction)

8. Something thrown through the air will have both a down (vertical) and across (horizontal) velocity • accelerate down (due to gravity) • go across at constant velocity(we assume no air resistance)

9. Each velocity will NOT affect the other (Independence) Vh (Vx) Vx is Constant Vy is changing (accelerating b/c of gravity) Vv (Vy)

11. What is the time of flight?What is the Vx?

12. Past D = Vt D= Vit + ½ at2 Vf= Vi + at Present X = Vxt Y = Vyi t+ ½ gt2 Y = ½ gt2 t2 = 2y/g Vyf = Vyi + gt Equations

13. Sample Problem • A stone is thrown horizontally at +15 m/s from the top of a cliff 44m high. • How long does the stone take to reach the bottom of the cliff? • How far from the base of the cliff does the stone strike the ground • Sketch the trajectory of the stone

14. 15 m/s Given Vx = 15m/s Vyi = 0 m/s G= -9.8m/s2 y = -44m 44m Formula X = Vxt Y = Vyt + ½ gt2

15. SOLUTION a. t2 = 2 y/g t2 = 2(-44)/ -9.8 t = √9 t= 3sec b. X= Vxt = (15 m/s) (3s) = 45m t= 3sec 44m 45m

16. Practice Problems • A stone is thrown horizontally at a speed of +5.0m/s from the top of a cliff 78.4m high. • How long does it take the stone to reach the bottom of the cliff? • How far from the base of the cliff does the stone strike the ground? • A steel ball rolls with a constant velocity across a tabletop .950m high. It rolls off and hits the ground +.352m horizontally from the edge of the table. How fast was the ball rolling?

19. Homework P. 102 1-3

20. Page 118 K and L

21. Objects Launched at Angles • When we analyze a trajectory, we should see what happens to vertical and horizontal components to understand the motion

22. Page 118 J • Find component velocities on the way up. • Find Resultant velocities on the way down.

23. Shallow Angle

24. Steep Angle

26. Use Vector Knowledge to Solve for Vx & Vy

27. The launching angle and the velocity determine how far the object will travel. • When the launching speed or force is the same, the angle alone will determine the range. • RANGE: how far (horizontally) a projectile goes

28. NOTE • For working with angled trajectories, the launching height is usually the same as the landing height When it is DY = 0!

29. Finding Time for angled launches • Time = 2Vy g Only works when landing/launching heights are the same. I call this the “hangtime” formula.

30. Example Problem for a projectile shot at an angle… • A small metal ball is shot with a velocity of 4.47m/s at an angle of 66° above the horizontal. • How long does it take the ball to land? • How high did the ball fly? • What was the range?

31. Solution • when land y=0 • 0 = Vyt + ½ gt2 • -Vyt = ½ gt2 • t = -2Vy/ g • t = -2(4.08)/-9.8 Formula y = Vyt + ½ gt2 X = Vxt Given Vi = 4.47m/s 0 = 66° 4.47 Vy 66 Vx Vy = 4.47(sin 66) = 4.08m/s Vx = 4.47(cos 66) = 1.82 m/s = .83 sec

32. B. max height =½t y = Vyt + ½ gt2 y = (4.08)(.417) + ½ (-9.8)(.417)2 y = 1.7 – 0.852 C. Flight time = .833sec X = Vxt = 1.82m/s (.833sec) y = 0.85m X = 1.52m

33. Practice problem 5 • A softball is thrown with an initial velocity of 27m/s at an angle of 30o from the horizontal. • A. Find the total time the ball is in the air. • B. find the horizontal displacement of the ball. • C. Find the maximum height for the ball.

34. 27 m/s Vy 30° Vx A. t = -2Vy/ g = -2(13.5)/- 9.8 = 2.76s B. X = Vxt = 23.38(2.76) =65.5m 5 Given Vx = 23.38 Vy = 13.5 g = -9.8m/s ½ t= 1.38 C. y = Vyt + ½ gt2 = 13.5(1.38) + ½ (-9.8)(1.38)2 18.63 + (-9.33) = 9.3m

35. Problem 6 • All the given information of this problem is the same as #5. The only difference is the launch angle is now 60o. • A. find time • B. Find range (X) • C. Find max height (Y).

36. 27 m/s Vy 60° Vx 6 A. t = -2Vy/ g -2(23.38)/-g = 4.78s B. X = Vxt 13.5(4.78) =64.5m Given Vx = 13.5 Vy = 23.38 g = -9.8 ½ t = 2.39s C. Y = Vyt+ ½ gt2 = 23.38 (2.39) + ½ (-9.8) 2.392 =27.9m

37. The Range Formula • A fast way to find range when initial velocity and angle are given (angle should be with the horizontal) Vi2 sin(2 0 ) R = g Can also Find Vi g R Vi2 = sin (2 0 )

38. Example • Find the range for an object shot at 50° with an initial velocity of 5m/s Solution = (5)2 sin (2x50°) / 9.8 + 24.62/ 9.8 = 2.51m Given Vi = 5m/s 0 = 50° R = ? g = 9.8 Formula R = Vi2 Sin 2 0 g

39. Solve Practice Problems p.104 3E #’s 5,2,1 Homework p.114-115 30, 32, 34, 36, 39

40. Complimentary Angles = same ranges 450 gives the longest range for a projectile shot without air resistance. While the ranges may be the same, the time in the air won’t.

41. What are satellites? • Natural • Artificial

43. How do satellites stay in orbit even though they have no propulsion? • Isaac Newton saw the answer to this by thinking about the Moon. He also knew over 400 years ago that it would be possible to orbit the Earth if mankind could go fast enough. He also predicted mankind would travel to the Moon.

45. For low Earth orbit(200 to 500miles high) • Tangential Velocity of 8000m/s (17,500mph) • Vt will change with altitude. • The object (satellite) will stay in orbit. • Every second it falls down 4.9m and moves downrange (tangentially) 8000m. The resultant path mirrors the curvature of the Earth.

46. Vectors and Projectiles Review Mr. Crabtree is standing on a perfectly good bridge in Zimbabwe when he decides to jump off. He is tied by his ankles to a bungee cord for safety. For 5.2 seconds, your teacher is in free fall. He jumped horizontally at 4 m/s as he left the bridge. The bridge is 150m above the Zambezi River. What was his vertical displacement at the end of the free fall? [neglect wind resistance]

48. Find horizontal displacement from the base of the building.

49. Find the time the ball is in the air 10m 20m