Motion in Two Dimensions

1. Motion in Two Dimensions LAB FOLDERS DUE TOMORROW!

2. (Ignore any effects from air resistance) A pickup is moving with a constant velocity and a hunter is sitting in the back pointing his rifle straight up. If he fires the rifle, where will the bullet land? In front of the pickup? Behind the pickup? Inside the pickup?

3. A rifle, at a height H above the ground, fires a bullet parallel to the ground. At the same instant and from the same height H, a second bullet is dropped from rest. Which bullet strikes the ground first?

4. While riding in a car traveling down the road, you reach out and drop a cat. What would its pathway to the ground look like to you? What would the cat’s pathway look like to a person standing alongside the road?

5. Projectiles Projectile: an object that moves through air or space, acted upon only by gravity Range: the horizontal distance traveled by a projectile Trajectory: the pathway of a projectile Gravity provides acceleration in the vertical direction. But….. Assuming no air resistance, there is NOTHING to affect its horizontal velocity so… The horizontal acceleration is ZERO

6. Drop an object or launch it horizontally? Which one hits the ground first? Gravity acts the same on both!

7. An object dropped from a given height will hit the ground AT THE SAME TIME As an object launched horizontally from the same height. Why? Gravity pulls both of them to the ground at the same rate. Acceleration due to gravity = -9.8 m/s/s

8. Projectiles Launched Horizontally d = vot + ½ at2 V = 20 m/s x (horizontal ) y (vertical) vo = 20 m/s vo = 0 a = 0 a = - g m/s2 Range (horizontal distance) = dx = votdy= Height = ½ gt2 H = ½ (9.8)t2 H = 50 m H = ½ (9.8)t2 Range (horizontal distance) = dx Range = 20 m/s x 3.19 s = 63.8 m

9. Projectiles Launched Horizontally V = 30 m/s d = vot + ½ at2 Height = ½(9.8)t2 Horizontal distance = dx = vot H = 50 m Range (horizontal distance) = dx Range = 30 m/s x 3.19 s = 95.7m

10. Projectiles Launched Horizontally V = 10 m/s H = 50 m Range (horizontal distance) = dx = vt Range = 10 m/s x 3.19 s = 31.9m

11. Projectiles Launched Horizontally V = ??? H = ½ gt2 Horizontal distance = dx = vt

13. Projectiles Launched Horizontally Time to ground = 3.19 seconds! H = 50 m

15. Think vertical Think horizontal A cannon ball is shot horizontally at 45 m/s off the top of a cliff 100 m high. How long did it take for the ball to hit the ground? How far did it land from the base of the cliff Height = ½ gt2 horizontal d = vot H ÷ ½(9.8) = t2 t = 4.52 s d = vot d = 45 m/s (4.52 s) = 203.4 m

16. Think vertical Think horizontal • A ball rolling at 8 m/s rolls off the top of a 20 m high building. How long will it take to hit the ground? How far will it land from the base of the building? Height = ½ gt2 horizontal d = vot H ÷ ½(9.8) = t2 t = 2.02 s d = vot d = 8 m/s (2.02 s) = 16.16 m

17. Think vertical Think horizontal • A projectile hits the ground 3 seconds after it was launched horizontally from the top of a hill. If the range of the projectile was 60 m, how fast was the projectile launched? How tall was the hill? Height = ½ gt2 horizontal d = vot H = ½(9.8) (3)2 = 44.1 m d = vot vo = d ÷ t = 60 ÷ 3 = 20 m/s

18. Think vertical Think horizontal • If a projectile hit the ground 4 seconds after it was launched and the range was 50 m, how fast was the launch velocity? How tall was the hill? Height = ½ gt2 horizontal d = vot H = ½ (9.8) (4)2 = 78.4 m d = vot vo = d ÷ t = 50 ÷ 4 = 12.5 m/s

19. Think vertical Think horizontal • A bullet is shot horizontally from the top of a hill. It strikes the ground in 1.5 seconds with a range of 700 m. How tall was the hill? How fast was the original velocity of the bullet? Height = ½ gt2 horizontal d = vot H = ½(9.8) (1.5)2 = 11.025 m d = vot vo = d ÷ t = 700 ÷ 1.5 = 466 m/s

20. LAB BOOKS DUE NOW!!

21. Projectiles Launched at an Angle The original velocity, vo of the projectile is not entirely horizontal nor entirely vertical, but has COMPONENTS that are both horizontal and vertical. vo vy vx

22. The vertical component of the velocity, vy is affected by gravity and will change at a rate of g = 10 m/s every second. (“g” is the magnitude of the acceleration due to gravity, but we understand the acceleration is in the downward direction) a = -g = -10 m/s2 The horizontal component of the velocity, vx is NOT affected by gravity and if there is no air resistance, it remains constant. vo vy vx

23. Vy = 50 m/s Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 1 second after the launch? Remember, a = - g = -10 m/s per second Vy = 40 m/s

24. Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 2 second after the launch? Vy = 30 m/s

25. Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 3 second after the launch? Vy = 20 m/s

26. Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 4 second after the launch? Vy = 10 m/s

27. Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 5 second after the launch? Vy = 0 m/s

28. Suppose the vertical component of the velocity, vy , was 50 m/s. What will it be 6 second after the launch? Vy = -10 m/s vo vx

29. Suppose the vertical component of the velocity, vy , was 50 m/s. How could one find the time it takes until a projectile reaches its highest point? vo Vy = 50 m/s vx

30. Suppose the vertical component of the velocity, vy , was 60 m/s. What will it be 1 second after the launch? 2 s? 3s? 4s? 5s? 6s? 7s? How could one find the time it takes until a projectile reaches its highest point? vo Vy = 60 m/s vx

31. Suppose the vertical component of the velocity, vy , was 55 m/s. What will it be 1 second after the launch? How could one find the time it takes until a projectile reaches its highest point? vo Vy = 55 m/s vx

32. Suppose the vertical component of the velocity, vy , was 40 m/s and the horizontal component of the velocity, vx, was 30 m/s. How long will it be until it reaches its highest point? 4s How much longer until it hits the ground? 4s How far in the horizontal x direction has the projectile traveled during those 8 seconds? In other words, what is its RANGE? Range = dx = vxt = (30 m/s) (8 s) Range = 240 m vo Vy = 40 m/s Vx = 30 m/s 240 m

33. Suppose the vertical component of the velocity, vy , was 60 m/s and the horizontal component of the velocity, vx, was 15 m/s. How long will it be until it reaches its highest point? 6s How much longer until it hits the ground? 6s How far in the horizontal x direction has the projectile traveled during those 12 seconds? In other words, what is its RANGE? dx = vxt = (15 m/s) (12 s) Range = 180 m vo Vy = 60 m/s Vx = 15 m/s 180 m

34. Suppose the vertical component of the velocity, vy , was 30 m/s and the horizontal component of the velocity, vx, was 50 m/s. How long will it be until it reaches its highest point? 3s How much longer until it hits the ground? 3s How far in the horizontal x direction has the projectile traveled during those 6 seconds? In other words, what is its RANGE? dx = vxt = (50 m/s) (6 s) Range = 300 m vo Vy = 30 m/s Vx = 50 m/s 300 m

35. Suppose the vertical component of the velocity, vy , was 45 m/s and the horizontal component of the velocity, vx, was 30 m/s. How long will it be until it reaches its highest point? 4.5s How much longer until it hits the ground? 4.5s How far in the horizontal x direction has the projectile traveled during those 9 seconds? In other words, what is its RANGE? dx = vxt = (30 m/s) (9 s) Range = 270 m vo Vy = 45 m/s Vx = 30 m/s 270 m

36. Suppose the vertical component of the velocity, vy , was 45 m/s and the horizontal component of the velocity, vx, was 30 m/s. How fast was vo? It’s not 45 m/s. It’s not 30 m/s. How would you find the original velocity, vo? Use the Pythagorean Theorem! vo Vy = 45 m/s Vx = 30 m/s

38. At what part of the trajectory does a projectile have minimum speed?

39. At what angle will a projectile go the farthest? 45 degrees

40. A ball is launched at an angle of 30 degrees and hits a target. At what other angle could the ball be launched and still hit the target? 45 – 30 = 15 degrees 45 + 15 = 60 degrees

41. If you throw a ball STRAIGHT up, what is its velocity at its highest point? ZERO What is its acceleration at its highest point? - 10 m/s2

42. If you throw a ball at an angle, what is its velocity at its highest point? Its velocity at its highest point is NOT zero, but is the same as its original x component of its velocity. What is its acceleration at its highest point? -10 m/s2

43. The following 2 slides are for Pre-AP only!

44. Analyzing projectiles using vector components Most of the time the horizontal and vertical components of vo are not provided and must be first identified by resolving the initial velocity vector. After identifying the components, the projectile can be analyzed in the same way for time to the highest point, range, etc.

45. *Note: Leave the velocity vector components in the form of vocosqor vosin q in your calculator rather than calculating out the value! This keeps errors to a minimum. Time to the highest point? = Time to ground? 0.7 s x 2 = 1.4 s Range?)t = (12cos35˚)1.4 = 13.76 m Max height? Since vf = 0 (in vertical direction) at the highest position, you can use vf2 = vo2 + 2ad to find max vertical height or… = 12 m/s vo vy = vosin q = 12 sin 35˚ q = 35 degrees vx = vocosq = 12 cos35˚

47. Two- Dimension Motion:Circular Motion

48. Question: Is there a constant velocity when an object moves in a circle with a constant speed? No, the direction changes, therefore the velocity changes. If the velocity changed, the object is actually ACCELERATING even while moving at the same speed. The change in direction is always pointed toward the center of the circle, so the acceleration is inward- toward the CENTER. “centripetal”- center seeking

49. Centripetal Acceleration acceleration acceleration acceleration

50. a EVERYTHING that moves in a circular path is experiencing a centripetal acceleration that points inward- toward the center of the circle!!