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Chapter 3

Chapter 3. Two-Dimensional Motion Projectiles launched at an angle. Some Variations of Projectile Motion. An object may be fired horizontally The initial velocity is all in the x-direction v i = v x and v y = 0 All the general rules of projectile motion apply.

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Chapter 3

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  1. Chapter 3 Two-Dimensional Motion Projectiles launched at an angle

  2. Some Variations of Projectile Motion • An object may be fired horizontally • The initial velocity is all in the x-direction • vi = vx and vy = 0 • All the general rules of projectile motion apply

  3. Projectile Motion at an angle

  4. How are they different? • Projectiles Launched Horizontally • The initial vertical velocity is 0. • The initial horizontal velocity is the total initial velocity. • Projectiles Launched At An Angle • Resolve the initial velocity into x and y components. • The initial vertical velocity is the y component. • The initial horizontal velocity is the x component.

  5. Some Details About the Rules • x-direction • ax = 0 • vx = vix = vicosΘi = constant • x = vixt • This is the only equation in the x-direction since there is constant velocity in that direction • Initial velocity still equals final velocity

  6. More Details About the Rules • y-direction • viy = visinΘi • Free fall problem • a = g • Object slows as it goes up (-9.8m/s2) • Uniformly accelerated motion, so the motion equations all hold • JUST LIKE STOMP ROCKETS • Symmetrical

  7. Problem-Solving Strategy • Resolve the initial velocity into x- and y-components • Treat the horizontal and vertical motions independently • Make a chart again, showing horizontal and vertical motion • Choose to investigate up or down. • Follow rules of kinematics equations

  8. Solving Launched Projectile Motion vi = (r , Θ) = (vix , viy) Horizontal Vertical a = vi = vf = t = x = 0 vix vix # # = range a = vi = vf = t = x = +/- 9.8m/s2 = g 0 or viy viy or 0 # Max height = y UP or DOWN INVESTIGATION… Where do the resolved components go?

  9. Projectile Motion at an angle

  10. Example 1: The punter for the Steelers punts the football with a velocity of 27 m/s at an angle of 30. Find the ball’s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.) • Looking for: • Total time (t) • Max height (y) • Range (x) Given: vi = (27m/s, 30o)

  11. 27m/s 30o What do we do with the given info? vi = (27m/s, 30o) What are the units? vi = (23.4m/s, 13.5m/s) “resolved” vector m/s Viy = 27sin30 Viy = 13.5 Viy = 13.5m/s Vix = 27cos30 Vix = 23.4 Vix = 23.4m/s

  12. So where does this info “fit” in the chart? Horizontal Vertical a = vi = vf = t = x = 0 23.4m/s 23.4m/s a = vi = vf = t = x = Viy if solving “up” = 13.5m/s Viy if solving “down” = 13.5m/s

  13. Pick a “side” to solve – symmetry Up: Horizontal Vertical a = vi = vf = t = x = 0 23.4m/s 23.4m/s a = vi = vf = t = y = - 9.8m/s2 13.5m/s 0 vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

  14. Projectile Motion at an angle

  15. Pick a “side” to solve – symmetry Down: Horizontal Vertical a = vi = vf = t = x = 0 23.4m/s 23.4m/s a = vi = vf = t = y = 9.8m/s2 0 13.5m/s vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

  16. Projectile Motion at an angle

  17. On the horizontal side • Max height occurs midway through the flight. • We found t = 1.38s both directions (up and down). • How long is the projectile in the air? DOUBLE this time for total air time t = 1.38 x 2 = 2.76s • What about range? x = vt x = (23.4)(2.76) x = 64.6m = RANGE

  18. This tells us… • Now we only need an initial velocity vector to determine all of the information we need to have a detailed description of where an object is in its path. vi = (r , Θ) = (vix , viy)

  19. Maximum Range vs. Maximum Height • What angle of a launched projectile gets the maximum height? • What angle of a launched projectile gets the maximum range? 90o 45o

  20. Projectile Motion at Various Initial Angles • Complementary values of the initial angle result in the same range • The heights will be different • The maximum range occurs at a projection angle of 45o

  21. Non-Symmetrical Projectile Motion • Follow the general rules for projectile motion • Break the y-direction into parts • up and down • symmetrical back to initial height and then the rest of the height

  22. your homework … • We are going to see what kind of job Hollywood writers and producers would do on their NECAP assessments… • Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.

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