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Projectile Motion

Projectile Motion. Projectile Motion. The path that a moving object follows is called its trajectory . Projectile motion involves the trajectories and velocities of any object that has been launched, shot, or thrown. Does this represent a realistic trajectory?. Yes. No. Maybe.

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Projectile Motion

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  1. Projectile Motion

  2. Projectile Motion • The path that a moving object follows is called its trajectory. • Projectile motioninvolves the trajectories and velocities of any object that has been launched, shot, or thrown.

  3. Does this represent a realistic trajectory? • Yes. • No. • Maybe.

  4. Does this represent a realistic trajectory?

  5. Does this represent a realistic trajectory? • Yes. • No. • Maybe. The coyote would not go straight horizontally, pause, and then fall straight down. You see unrealistic trajectories all the time in media. Can you think of any others?

  6. Rules There are only a few rules we have to follow: All projectiles are freefalling vertically with an acceleration of 9.8 m/s2downwards Horizontal motion is totally unaffected by gravity! Since there are no forces acting on it, a falling object’s horizontal velocity is constant!

  7. Visualizing Projectiles • first enter vectors • focus on vx vx is constant the whole flight!

  8. Visualizing Projectiles no vy at the top! • first enter vectors • focus on vx • focus on vy vy decreases as it rises! by how much per second?

  9. Visualizing Projectiles

  10. 0

  11. Horizontal Projectiles • Horizontal motion is constant – there is no acceleration • Only formula used in horizontal (x) direction is: vx = dx/ t constant speed!

  12. Horizontal Projectiles • Horizontal projectiles are not thrown up or down. They are moving horizontally and falling vertically • The onlyinitial velocity is in the x direction • Vertical velocity (vy) is gained by freefall • viy = 0 • since viyis in freefall, • a = -9.8 m/s2

  13. HORIZONTAL PROJECTILE pg 288 • A rock is thrown horizontally from the top of a cliff at a constant speed of 15 m/s. Calculate the horizontal& verticalpositionsof the rock after each second and place these positions in the table below. Assume the rock is freefalling from rest.

  14. ANGLED PROJECTILE pg 289 A rock is thrown at an angle of 30° from the top of a cliff. Calculate the horizontal& verticalpositionsof the rock after each second and place these positions in the table below.

  15. Let’s analyze the jump dmax

  16. Let’s analyze the jump

  17. http://www.physicsclassroom.com/mmedia/vectors/pap.cfm

  18. Variable Definitions: vyi initial velocity in y direction vx= constant velocity in x direction v final velocity of projectile dx  horizontal range dy  Height dy Maximum altitude dy Vertical displacement

  19. Hitting a Target • If the rifle is fired directly at the target in a horizontal direction, will the bullet hit the center of the target? • Does the bullet fall during its flight?

  20. ExampleA person decides to fire a rifle horizontally at a bull’s-eye. The speed of the bullet as it leaves the barrel of the gun is 890 m/s. He’s new to the ideas of projectile motion so doesn’t aim high and the bullet strikes the target 1.7 cm below the center of the bull’s-eye. What is the horizontal distance between the rifle and the bull’ s-eye? start by drawing a picture: label the explicit givens with subscript. Ex: dxor dy

  21. ExampleWhat is the horizontal distance between the rifle and the bull’ s-eye? givens (separate by direction): unknown: horizontaldisplacement

  22. Example which equation do we use? use to find time rewrite equation for t

  23. Example Use t and vx to solve for dx

  24. Non-horizontal Projectiles • vx is still constant • vy is still in freefall • only difference with non-horizontal is… • now the object begins with a vertical velocity!

  25. Non-horizontal Projectiles • Angled Projectiles require a little work to get useful vi • vi has an x and y component • need to calculate initial vx and vy

  26. Breaking up a vector • every vector has 2 components to it • a horizontal component • a vertical component • they add up to the total

  27. Breaking up a vector SOHCAHTOA hypotenuse opposite adjacent

  28. Breaking up a vector SOHCAHTOA need to find θ? hypotenuse opposite adjacent

  29. Non-horizontal Projectiles • need to calculate initial vx and vy

  30. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vi=31.5 m/s 1. Start by drawing a picture: vyi 2. Find the initial velocity in the x and y directions vx

  31. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vi=31.5 m/s Decide upon initial and final conditions vyi vx

  32. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vi=31.5 m/s There are 2 ways to solve vyi Find the velocity when the ball hits the ground, then find the time it takes to get there. Use the quadratic formula vx

  33. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vi=31.5 m/s Equations vyi vx a b c Solve using Quadratic formula … or GRAPH it!

  34. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? When is the stone on the ground? vo = 31.5m/s vyo vx t = 7.00s

  35. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What is the range of the stone? vi=31.5 m/s vyo vx Equations

  36. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? vx vyf vf= ?

  37. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? vx vyf vf= ?

  38. Angled Projectile ExampleA stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? Givens Equations vx vyf vf= ?

  39. Varied Angles • which projectile angle shoots highest? • larger θmeans faster viy • which projectile angle shoots farthest? • 45° has perfect balance of fast vxand long flight time.

  40. Practice pg 298

  41. 1.5 m above the 6. When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump ground 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide what is the maximum number of friends over which the cyclist can jump? horizontal. v = 14 m/s

  42. When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? 1. Find t which equation do we use? We could use to find time But you would have to use the quadratic formula to solve for t because there is a t and t2

  43. When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? 1. Find t which equation do we use? Instead, find vyf vyf = - 6.9 m/s And then find t: t = 1.14 s

  44. When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? 2. Find dx vyf = - 6.9 m/s t = 1.14 s 3. How many friends?

  45. Varied Angles • which projectile angle shoots highest? • larger θmeans faster viy • which projectile angle shoots farthest? • 45° has perfect balance of fast vxand long flight time.

  46. ` For the following situations: State if the following are positive, negative or zero. vo v a x

  47. 3. Calculate the maximum altitude of the ball (from the floor). ? vy = vfy 0 m/s vfy

  48. 3. Calculate the maximum altitude of the ball (from the floor). 1. Find t which equation do we use? ? vy = 0 m/s vfy= from #2

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