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CHAPTER 3 PROJECTILE MOTION

CHAPTER 3 PROJECTILE MOTION. VECTORS. North. positive y. West. East. negative x. positive x. negative y. South. X-component 10 . Y-component 0. VECTOR EXAMPLE 1. 10 m East. 10 m. VECTOR EXAMPLE 2. 10 m West. 10 m. X-component -10 . Y-component 0. VECTOR EXAMPLE 3.

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CHAPTER 3 PROJECTILE MOTION

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  1. CHAPTER 3PROJECTILE MOTION

  2. VECTORS North positive y West East negative x positive x negative y South

  3. X-component 10 Y-component 0 VECTOR EXAMPLE 1 10 m East 10 m

  4. VECTOR EXAMPLE 2 10 m West 10 m • X-component -10 • Y-component 0

  5. VECTOR EXAMPLE 3 10 m North 10 m • X-component 0 • Y-component 10

  6. VECTOR EXAMPLE 4 10 m South 10 m • X-component 0 • Y-component -10

  7. COMPONENTS OF A VECTOR y VY V  x VX

  8. EXAMPLE A wind with a velocity of 40 m/s blows towards 30 NE. What are the x and y components of the wind’s velocity. V=40 Vy  Vx

  9. PROJECTILE A projectile is an object with an initial velocity that is allowed to move under the affects of gravity. Ex. a javelin throw, a package released by an airplane, a thrown baseball

  10. Types of Projectiles

  11. Because of Earth’s Gravitational pull and their own inertia, projectiles follow a curved path. Trajectory: is the path of the projectile. They have both horizontal and vertical velocities.

  12. Constant horizontal velocity due to inertia Vix = Vfx ax = 0

  13. Constant horizontal velocity due to inertia ViY = 0 aY = -9.8 m/s2

  14. IMPORTANT!!! The ball’s horizontal and vertical motions are completely independent of each other.

  15. 3 1 2 CURVED MOTION UNDER GRAVITY 1. Ball is simply dropped 2. Ball is thrown horizontally at 10 m/s 3. Ball is thrown horizontally at 20 m/s

  16. In the absence of any frictional forces (like air), all the three balls fall to the ground at the same time. The horizontal motion does not affect the vertical acceleration. All the three balls are pulled to the ground in the same way because of gravity. All the three balls travel the same vertical distance in the same time.

  17. PROJECTILES LAUNCHED HORIZONTALLY INITIAL VELOCITY TRAJECTORY HEIGHT (DY) RANGE (DX)

  18. X-component initial velocity horizontal distance (range) zero acceleration final velocity= initial velocity Y- component zero initial velocity negative vertical distance -9.8 m/s2 acceleration Negative final velocity

  19. PROBLEM 1 A stone is thrown horizontally at 15 m/s from the top of a cliff 44 m high. A) How long does the stone take to reach the ground? B) How far from the base of the cliff does the stone strike the ground? C) Sketch the trajectory of the stone.

  20. X-component Vi = 15 m/s a = 0 m/s2 Vf =15 m/s Y- component Vi = 0 m/s a = -9.8 m/s2 D = - 44 m

  21. X-component Vi = 15 m/s a = 0 m/s2 Vf =15 m/s Y- component Vi = 0 m/s a = -9.8 m/s2 D = - 44 m D = Vit + ½ at2 t = 3s D = Vit + ½ at2 D = 15(3) = 45m - 44 = ½ (-9.8)t2 t = 3 s

  22. PROBLEM 2 A stone is thrown horizontally at a speed of + 5 m/s from the top of a cliff 78.4 m high. A) How long does the stone take to reach the bottom of the cliff? B) How far from the base of the cliff does the stone strike the ground? C) What are the horizontal and vertical velocities of the stone just before it hits the ground?

  23. X-component Vi = 5 m/s a = 0 m/s2 Vf =5 m/s Y- component Vi = 0 m/s a = -9.8 m/s2 D = - 78.4 m

  24. X-component Vi = 5 m/s a = 0 m/s2 Vf =5 m/s Y- component Vi = 0 m/s a = -9.8 m/s2 D = - 78.4 m D = Vit + ½ at2 t = 4s D = Vit + ½ at2 D = 5(4) = 20 m -78.4=½ (-9.8) t2 t = 4 s

  25. Y- component Vi = 0 m/s a = -9.8 m/s2 D = - 78.4 m t = 4 s Vf = Vi + at = 0+(-9.8) 4 =- 39.2 m/s

  26. PROBLEM 3 How would a) and b) change if the stone is thrown with twice the velocity. Answer: A) would not change B) would increase two times

  27. PROBLEM 4 How would a) and b) change if the stone is thrown with the same speed but twice the height. Answer: A) would increase B) would increase

  28. PROBLEM 5 A steel ball rolls with a constant velocity across a table top 0.950 m high. It rolls and hits off the ground + .350 m horizontally from the edge of the table. How fast was the ball rolling?

  29. X-component Vi = ? m/s a = 0 m/s2 Vf =? m/s D = +.350 m Y- component Vi = 0 m/s a = -9.8 m/s2 D = - .950 m

  30. X-component Vi = ? m/s = Vf a = 0 m/s2 D = .350 m t = .44 s D = Vit + ½ at2 0.350 = Vi (.44) Vi = .80 m/s Y- component Vi = 0 m/s a = -9.8 m/s2 D = -0.950 m D = Vit + ½ at2 -0.950= ½(-9.8)t2 t = .44 s

  31. Suppose you throw a ball upward at an angle. The ball has two types of motion a) Horizontal Motion b) Vertical Motion

  32. The ball moves horizontally at a constant speed. As it moves up gravity slows the ball down (the y speed decreases). At the maximum height, the ball stops. It changes direction and falls downward. Gravity increases the speed of the ball as it falls.

  33. The path of motion is an arc-shaped curve known as a parabola. Horizontal motion has no effect on the time it takes an object to fall to the ground.

  34. PROJECTILES LAUNCHED AT AN ANGLE LAUNCHING VELOCITY Y-COMP LAUNCHING VELOCITY  LAUNCHING VELOCITY X-COMP

  35. X-component ViX = V cos  horizontal distance (range) zero acceleration final velocity= initial velocity Y- component ViY = V sin  @ maximum height (y velocity = 0) -9.8 m/s2 acceleration At the end. D=0 Final Velocity=Negative LAUNCHING VELOCITY = v

  36. PROBLEM 6 The initial velocity of a ball in projectile motion is 4.47 m/s. It is projected at an angle of 66 above the horizontal. Find A) how long did it take to land.? B) how high did the ball fly? C) what was its range?

  37. X-component ViX=4.47 cos 660 = 1.8 m/s aX = 0 m/s2 VfX =1.8 m/s Y- component ViY=4.47 sin 660 =4.1 m/s aY =-9.8 m/s2 4.47 m/s B 660 A C

  38. X-component ViX=4.47 cos 66 = 1.8 m/s aX = 0 m/s2 VfX =1.8 m/s Y- component ViY=4.47 sin 66 = 4.1 m/s aY = -9.8 m/s2 At B VfY = 0 m/s 2aY DY = VfY2 - ViY2 DY = .86 m At B (Y-comp) VfY = ViY + aY t 0=4.1+(-9.8)t t = .417 s

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