1 / 32

Vectors and Projectile Motion

Vectors and Projectile Motion. Chapter 3. Adding Vectors. When adding vectors that fall on the same line, using pluses and minuses is sufficient.

hong
Download Presentation

Vectors and 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. Vectors and Projectile Motion Chapter 3

  2. Adding Vectors • When adding vectors that fall on the same line, using pluses and minuses is sufficient. • When dealing with two dimensions, vector addition must be used. A complete answer includes magnitude, units, and a direction usually indicated by an angle.

  3. Right Angle Vector Addition • C=resultant, means sum of two vectors. • In order to identify the angle, vectors must be lined up head to tail. 4m 2m • In this case, Pythagorean theorem will solve for c. q =63.4o c =4.47m • In order to find , trig must be used. It is usually best to use the given sides, rather than the calculated side. If you walk 2m north and then 4m east, what is your displacement?

  4. 5m/s 10m/s Practice Problem A sailboat experiences a 5m/s wind to the east while traveling 10m/s downstream. Find the resulting net velocity 5m/s 10m/s

  5. Adding more than two vectors • Vectors can be added in any order. • Subtracting a vector is the same as adding a vector that is the same length, but going the exact opposite direction. • When drawing a picture, vectors may be rearranged into any order as long as their orientation is maintained.

  6. Example

  7. Practice • What is the total displacement of the motion, assuming each block is 1m long?

  8. Vector direction • Degrees on the unit circle • Start on the positive x and move CCW for positive degrees and CW for negative degrees. • Examples, 40m @ 217°, 34m/s @ -73° • Compass direction • 27° S of W • Southwest would indicate a 45° • Earth and space sciences use a system where north is 0° and one always has a positive angle and proceeds CW around the compass.

  9. Adding vectors with random orientation • Components-A set of vectors that have a sum equal to a given vector. • Usually most useful to have components perpendicular to each other. • Breaking up a vector, called resolution, is the exact opposite of adding two perpendicular vectors to find a resultant. • Usually, components are along the x and y axis. x y

  10. Component practice 32m@67° 50m@200° 14m/s@-112° 6m/s@-300°

  11. Unit Vector notation • In some areas of math and science, vectors are frequently given according to their components rather than their resultants. • Each vector is therefore listed as a pair of vectors, with the length of each vector stated numerically and the direction of the vector indicated by a unit vector, or • The unit vector is a vector with a length of 1 in either the x axis direction or the y axis direction. • Ex: 10m@43° can be written as

  12. Steps for solving vector problems • Graphically represent the system to give yourself a check for your answer. • Draw each vector individually. • Using trig, find the x and y components of each vector. • Add all the x components together and add all the y components together. • Make a new diagram using the total x component and the total y component. Draw in the resultant and label a reference angle. • Use the Pythagorean theorem and trig to find the resultant and angle

  13. Example Problem 9m, 45o up from the horizontal 11m, 18o down from the horizontal

  14. Splitting the Triangles Up 11m, 18o down from the horizontal 9m, 45o up from the horizontal x2 18o y1 y2 45o x1

  15. x2 18o y1 y2 45o x1

  16. 9m, 45o up from the horizontal 45o x1 Working with the first triangle y1

  17. Working with the second triangle 11m, 18o down from the horizontal x2 18o y2

  18. c ytot q xtot Putting everything together

  19. Practice • Find the resultant of the following vectors: • 450m @ 20o • 360m @ 300o • 290m @ 189o • 405m @ 115o

  20. Projectile Motion

  21. Horizontal and Vertical Motion

  22. Things to remember for 2D motion • Vertical and horizontal motion are independent. • When something is traveling through the air, ignore the effects of air resistance. • There is nothing pushing or pulling a projectile horizontally, therefore ax=0. • For vertical motion, gravity is causing the vertical acceleration, so ay=-9.8m/s2. • We will assume that projectiles landing at a height different from their initial height are always launched horizontally. Therefore, any initial velocity is an x piece. There is no y component for initial velocity.

  23. Cliff Problems • A car drives off a 100m cliff at a speed of 47m/s. What is: • The time it takes to hit the ground? • It’s horizontal distance from the base of the cliff? • It’s final velocity?

  24. Horizontal vi=47m/s vf=47m/s a=0 t=? x=? Vertical a=g=-9.8m/s2 vi=0 y=100m t=? vf=? Identify Vertical and Horizontal Components

  25. Part A • Use y pieces because there is not enough information to solve for t using x components. • Since the car stops moving horizontally at the same time it stops moving vertically, the t found using the y components can be used for the x components as well.

  26. Part B

  27. Part C • The car is going both down and over at the end so vf has both x and y components. • vfx= vix • vfy must be calculated.

  28. vfx q c vfy Part C Continued • Put x and y components together and solve for both resultant and angle.

  29. A more complicated example • A water balloon is launched out of a slingshot from a height of 1.7m at a window which is 7m above the ground. Assuming the launcher is aimed at a 56° angle and the window and the launcher are 10m apart, what is the launch speed needed for the balloon to make it to the window?

  30. Practice Problem • An onion runs off of a building at a speed of 22.2m/s and lands 36.0m from the base of the building. How tall is the building? • A ball is thrown from the roof of a building 56m tall and lands 45m away from the base. What was the ball’s launch angle assuming it’s initial speed was 20m/s?

  31. Range equation • Range-the horizontal distance travelled by a projectile. • Equation can only be used if the object following a parabolic path is starting and ending at the same height. • Derivation can be found on page 88 of your text.

  32. Practice Problem • A water balloon is shot from a slingshot at an angle of 21o and with a velocity of 16m/s. What is: • The horizontal range? • The time it’s in the air? • It’s maximum height?

More Related