1 / 35

PROJECTILE MOTION

PROJECTILE MOTION. 2-D MOTION. Projectile Motion. Curve path or trajectory x and y motion at the same time Use the 4 kinematics equations x direction: x, v, t y direction: y, v f , v o , a, t 2 types Launched horizontally Launched at an angle. Launched Horizontally. x-direction NO a

stocking
Download Presentation

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. PROJECTILE MOTION 2-D MOTION

  2. Projectile Motion • Curve path or trajectory • x and y motion at the same time • Use the 4 kinematics equations • x direction: x, v, t • y direction: y, vf, vo, a, t • 2 types • Launched horizontally • Launched at an angle

  3. Launched Horizontally • x-direction • NO a • vº = vf • y-direction • Δ uniformly • a = g • vº = 0 m/s • ty = tx

  4. Launched Horizontally

  5. Launched Horizontally

  6. Example 1 • A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. • How long does it take the stone to reach the bottom of the cliff? • How far from the base of the cliff does the stone strike the ground? • What are the horizontal and vertical components of the velocity of the stone just before it hits the ground?

  7. Example 1 • Draw Diagram • Make a table

  8. Example 1 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(78.4 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 4.0 s

  9. Example 1 • Solve for x • x = vxt = (5.0 m/s) (4.0 s) = 20. m

  10. Example 1 • Solve for vx and vy final • vx is constant = 5.0 m/s • vy is changing • vy = vº+ gt = gt = (9.8m/s/s) (4.0 s) = 39 m/s

  11. Example 2 • A steel ball rolls with constant velocity across a tabletop 0.950 m high. It rolls off and hits the ground 0.352 m horizontally from the edge of the table. How fast was the ball rolling?

  12. Example 2 • Draw Diagram • Make a table

  13. Example 2 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(0.950 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 0.440 s

  14. Example 2 • Solve for vx • x vx = ------ t • 0.352 m vx = ------------- 0.440 s = 0.800 m/s

  15. Example 3 • An auto, moving too fast on a horizontal stretch of mountain road, slides off the road, falling into deep snow 43.9 m below the road and 87.7 m beyond the edge of the road. • How long did the auto take to fall? • How fast was it going when it left the road? • What was its acceleration 10 m below the edge of the road?

  16. Example 3 • Draw Diagram • Make a table

  17. Example 3 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s

  18. Example 3 • Solve for vx • x vx = ------ t • 87.7 m vx = ------------- 2.99 s = 29.3 m/s

  19. Example 3 Acceleration is only due to gravity.

  20. Launched at an Angle • x-direction • NO a • vº constant • y-direction • Δ uniformly • a = g • tup = tdown • ty = tx • Solve x and y

  21. Example 4 • A player kicks a football from ground level with a velocity of magnitude of 27.0 m/s at an angle of 30.0º above the horizontal. • Find the hang time. • Find the distance the ball travels before it hits the ground. • Find its maximum height.

  22. Example 4 • Draw Diagram • Make a table

  23. Launched at Angle • Solve for initial vx and vy • vx = v cosθ = (27 m/s) cos 30º = 23 m/s • vy = v sinθ = (27 m/s) sin 30º = 14 m/s

  24. Example 4 • Solve for time • vy = vo + gt - 14 m/s = 14 m/s + (-9.8 m/s2) t -28 m/s = -9.8 m/s2 t t = 2.8s

  25. Example 4 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s

  26. Example 4 • Solve for time • y = vot + ½gt2 = ½gt2 • 2y 2(43.9 m) t = -------- = -------------- √ g √ 9.8 m/s/s = 2.99 s

  27. Example 5 • A rude tourist throws a peach pit horizontally with a 7.0 m/s velocity out of an elevator cage. • If the elevator is not moving, how long will the pit take to reach the ground, 17.0 m below? • How far from the elevator will the pit land? • He throws the next pit when the elevator is at the same height but moving upward at a constant 8.5 m/s velocity. How long will it take this pit to land? • How far away will this pit land?

  28. Example 5 • Draw Diagram • Make a table

  29. Graphing Projectile Motion Height Horizontal Distance

  30. Graphing Projectile Motion Height Time

  31. Graphing Projectile Motion Vertical Speed Time

  32. Graphing Projectile Motion Horizontal Velocity Time

  33. Graphing Projectile Motion Horizontal Distance Time

More Related