1 / 135

Chapter 5 Projectile motion

Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion. Chapter 5: considers motion that follows a diagonal path or a curved path. When you throw a baseball, the trajectory is a curved path.

caldwell-morton
Download Presentation

Chapter 5 Projectile motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5 Projectile motion

  2. Chapter 4: straight line motionthat was ONLY vertical or ONLY horizontal motion

  3. Chapter 5: considers motion that follows a diagonal path or a curved path

  4. When you throw a baseball,the trajectory is a curved path.

  5. We are going to separate the motion of a projectile into independent x and y motions

  6. The vertical motion is not affected by the horizontal motion. And the horizontal motion is not affected by the vertical motion.

  7. Observe: a large ball bearing is dropped at the same time as a second ball bearing is fired horizontally.

  8. What happened?

  9. Remember adding 2 perpendicular vectorshorizontal and vertical vectors.

  10. When we add perpendicular vectors we use Pythagorean theorem to find the resultant.

  11. Consider a vector B that is pointed at an angle q wrt horizontal direction.

  12. We are going to break vector B into 2 perpendicular vectors: Bx and By

  13. if you ADD vectors Bx + By you get vector B.

  14. Graphically, we can say:

  15. Draw a rectangle with vector B as the diagonal. the component vectorsBx and By are the sides of the rectangle

  16. Application:

  17. A Boat in a river

  18. How can we describe the motion of a boat in a river?

  19. The motion is affected by the motor of the boat and by the current of the river

  20. Imagine a river 120 meters wide with a current of 8 m/sec.

  21. Imagine a river 120 meters wide with a current of 8 m/sec.

  22. If a boat is placed in the river [motor is off] , how fast will the boat drift downstream?

  23. If the boat is drifting, the total speed of the boat just equals the speed of the current.

  24. for a boat drifting with the current: Vtotal= Vboat + VcurrentVtotal = -0 + Vcurrent = 8 m/sec

  25. Now suppose this boat can travel at a constant 15 m/sec when the motor is on .

  26. What is the total speed of the boat downstream when the motor is on?

  27. The boat is traveling in the same direction as the current.

  28. V total downstream = Vtotal = Vboat + VcurrentVtotal = 15↓ + 8↓ = 23 m/sec ↓

  29. What is the total speed of the boat traveling upstream[against the current] ?

  30. The boat and current now move in opposite directions

  31. Vtotal = Vboat +VcurrentVtotal = 15 ↓+ ( 8 ↑) Vtotal = 7↓ m/sec

  32. Summary:traveling downstream: Vboat + VcurrentTraveling upstream:Vboat + [-Vcurrent]

  33. Crossing the river.

  34. If there was no current, how many seconds needed for this boat to travel 120 meters from A to B?

  35. Velocity = distance timeso time = distance velocity

  36. time = distance velocitytime = 120 m 15 m/sectime = 8 seconds

  37. But there if IS a current. what happens when you try to go straight across the river from A to B?

  38. The boat will travel from A to C.

  39. Every second the boat travels ACROSS 15 meters and AT THE SAME TIMEevery secondthe boat will be pushedDOWNSTREAm 8 meters by the current .

  40. Vboat = 15  and Vcurrent = 8↓These velocities are perpendicular

  41. The RESULTANT velocity of the boat is

  42. Vresultant2 = Vboat2 + Vcurrent2Vresultant2 = 152+ 82Vresultant2 = 225+ 64 =289Vresultant = 17 m/sec

  43. The boat still crosses the river in 8 seconds , but it lands downstream at point C not at point B.

More Related