Download
chapter 5 projectile motion n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 5 Projectile motion PowerPoint Presentation
Download Presentation
Chapter 5 Projectile motion

Chapter 5 Projectile motion

233 Views Download Presentation
Download Presentation

Chapter 5 Projectile motion

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chapter 5 Projectile motion

  2. 1. Recall: a projectile is an object only acted upon by gravity

  3. 2.Chapter 4: [linear motion]straight line motionthat was ONLY vertical or ONLY horizontal motion

  4. 3. Chapter 5: looks at motion that followsa diagonal path or a curved path

  5. 4. When you throw a baseball, it travels in an curved path.

  6. 5. We will separate curved motion into independent x and y motions

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

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

  9. What happens?

  10. Remember adding 2 perpendicular vectorshorizontal and vertical vectors.

  11. 8. Remember: When we add perpendicular vectors we use Pythagorean theorem to find the resultant.

  12. Boat in a river

  13. 9. a river is 120 meters wide and has a current of 8 m/sec.

  14. 9. a river is 120 meters wide and has a current of 8 m/sec.

  15. Traveling up and down stream

  16. 10. How fast will a boat drift downstream?

  17. 11. Vtotal = Vboat + VcurrentVtotal = 0 + 8 = 8 m/sec

  18. Now using the motor…

  19. 12. Suppose the motor moves the boat at 15 m/sec.

  20. 12. Suppose the motor moves the boat at 15 m/sec.how fast will the boat travel downstream [with the current]?

  21. 13. Using motor with current: Vtotal= Vboat + Vcurrent

  22. Total velocity traveling downstream: Vtotal= Vboat + Vcurrent= 15 + 8 = 23 m/sec 

  23. 14.Using motor against the current: total velocity traveling upstream [AGAINST the current] Vtotal = Vboat –Vcurrent

  24. 15. Going upstream: Velocity of current and boat are opposite directionsVtotal = Vboat –Vcurrent

  25. Vtotal = 15 -8 = 7 m/sec 

  26. Crossing the river

  27. 16. If there was no current, how many seconds for the boat to travel 120 meters to reach the opposite side?

  28. Velocity = distance timeor time = distance velocity

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

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

  31. 18. If there is a current, The boat still crosses in 8 seconds, but it lands downstream at point C

  32. 19. Add the perpendicular velocity vectors add to find the resultant velocity

  33. The triangles are similar:

  34. 20. In this example, Every second the boat travels 15 meter in the x direction IT ALSO TRAVELS 8 meter in the y direction

  35. How far down stream is the boat when it reaches the opposite shore?

  36. Velocity = distance timeso distance = velocity X timedistance = 8 m/sec X 8 sec = 64 m

  37. What if you want to travel straight across and land at B, not C?

  38. 21) If you want to go from A to B instead, you must point the boat diagonally upstream to compensate for the current.

  39. 22) Planes are affected by the wind the same way

  40. 23) Head wind: slows the plane[opposite direction]

  41. 24. Tail wind: speeds the plane up [same direction]

  42. 25. crosswind: blows plane off course [wind perpendicular to direction of plane]

  43. Break Vboat into VxandV y components

  44. Use pythagorean theorem to find Vx.

  45. Vx2 + Vy2 = Vboat2Vx2 + 82 = 152Vx2 =225-64 = 161Vx =12.7 m/sec

  46. How many seconds to cross?Velocity = Distance/timeT = D/V

  47. How many seconds to cross? From A to BT = D/VT = 120/12.7 T = 9.4 sec