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

3. Motion for a Projectile is described in terms of… Position Position Velocity Velocity Acceleration Acceleration All three of these are vector quantities and will be represented with arrows like the previous topic All three of these are vector quantities and will be represented with arrows like the previous topic We must remember that horizontal and vertical velocities (components) of a projectile are INDEPENDENT! We must remember that horizontal and vertical velocities (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 Something thrown through the air will have both a down (vertical) and across (horizontal) velocity 1. accelerate down (due to gravity) 2. go across at constant velocity (we assume no air resistance)

9. Each velocity will NOT affect the Each velocity will NOT affect the other (Independence) V h (V x ) V v (V y ) V x is Constant V y is changing (accelerating b/c of gravity)

11. What is the time of flight? What is the V x ?

12. Equations Past D = Vt D= V i t + ½ at 2 V f = V i + at Present X = V x t Y = V yi t+ ½ gt 2 Y = ½ gt 2 t 2 = 2y / g V yf = V yi + gt

13. Sample Problem A stone is thrown horizontally at +15 m/s from the top of a cliff 44m high. A stone is thrown horizontally at +15 m/s from the top of a cliff 44m high. 1. How long does the stone take to reach the bottom of the cliff? 2. How far from the base of the cliff does the stone strike the ground 3. Sketch the trajectory of the stone

14. 44m 15 m/s Given V x = 15m/s V yi = 0 m/s G= -9.8m/s 2 y = -44m Formula X = V x t Y = V y t + ½ gt 2

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

16. Practice Problems 1. A stone is thrown horizontally at a speed of +5.0 m / s from the top of a cliff 78.4m high. a. How long does it take the stone to reach the bottom of the cliff? b. How far from the base of the cliff does the stone strike the ground? 2. 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 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 V x & V y

27. The launching angle and the velocity determine how far the object will travel. 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. When the launching speed or force is the same, the angle alone will determine the range. RANGE: how far (horizontally) a projectile goes 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  Y = 0! For working with angled trajectories, the launching height is usually the same as the landing height When it is  Y = 0!

29. Finding Time for angled launches Time = 2V y 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 4.47 m / s at an angle of 66° above the horizontal. A small metal ball is shot with a velocity of 4.47 m / s at an angle of 66° above the horizontal. a. How long does it take the ball to land? b. How high did the ball fly? c. What was the range?

31. Given V i = 4.47 m / s 0 = 66° VxVxVxVx VyVyVyVy 4.47 66 Formula y = V y t + ½ gt 2 X = V x t Solution A.when land y=0 0 = V y t + ½ gt 2 -V y t = ½ gt 2 t = -2V y / g t = -2(4.08) / -9.8 V y = 4.47(sin 66) = 4.08 m / s V x = 4.47(cos 66) = 1.82 m / s =.83 sec

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

33. Practice problem 5 A softball is thrown with an initial velocity of 27m/s at an angle of 30 o 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. Given V x = 23.38 V y = 13.5 g = -9.8m/s ½ t= 1.38 yVyyVy 30° VxVxVxVx 27 m/s A. t = -2V y / g = -2(13.5)/- 9.8 = 2.76s B. X = V x t = 23.38(2.76) =65.5m C. y = V y t + ½ gt 2 = 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 60 o. A. find time B. Find range (X) C. Find max height (Y).

36. yVyyVy VxVxVxVx 27 m/s 60° Given V x = 13.5 V y = 23.38 g = -9.8 ½ t = 2.39s A. t = -2V y / g -2(23.38)/-g = 4.78s B. X = V x t 13.5(4.78)=64.5m C. Y = V y t+ ½ gt 2 = 23.38 (2.39) + ½ (-9.8) 2.39 2 =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) A fast way to find range when initial velocity and angle are given (angle should be with the horizontal) R = V i 2 sin(2 0 ) g Vi2Vi2 g R sin (2 0 ) = Can also Find V i

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

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

40. Complimentary Angles = same ranges 45 0 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) V t 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