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PROJECTILES

Two -Dimentional Motion. PROJECTILES. Projectile Motion. Objects thrown or launched into the air and subject to gravity. Ball A is dropped from the edge of a table while Ball B is simultaneously rolled off the edge with the initial speed V i. V i. B. A. A. B.

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PROJECTILES

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  1. Two -Dimentional Motion PROJECTILES

  2. Projectile Motion Objects thrown or launched into the air and subject to gravity.

  3. Ball A is dropped from the edge of a table while Ball B is simultaneously rolled off the edge with the initial speed Vi Vi B A A B

  4. Ball A has horizontal AND vertical velocity • Acceleration of gravity is the same for both • balls regardless of their state of motion. • Both take the same time to reach the ground Vi B A t =1s t =1s A B

  5. Resolving Vectors in 2 Dimensional MotionTo analyze projectile motion we must resolve (separate) its motion path into a horizontal x component and vertical y component.

  6. The combined horizontal component (x) and vertical component (y) of an objects motion IS the RESULTANT vector. The resultant vector is the path we see the object going.

  7. As a ball rolls off the table it has a horizontalforward initial velocity off the table.The initial velocity of the ball is constant all the way down.Verticallythe ball is in freefall downward. Vi

  8. Apply the horizontal and vertical components to the path as it falls. • Initial horizontal velocity of ball Vix = constant • Final velocity of ball when it reaches the ground is • Vyf = - gt • Vyf =- 9.81 m/s2 x 0.45 s = 4.42 m/s 5 m/s B

  9. If we resolve the motion of the ball into its horizontal component V xiAND vertical component Vyi V xi Rv Vyi

  10. Vix = constant Vfy= gt dX= VixtVfy2 = 2 gy dy= ½ gt2 acceleration of gravity is negative -9.8 m/s Horizontally Launched Projectile Equations Horizontal MotionVertical Motion

  11. Problem A ball rolls off a table with an initial velocity of 5 m/s. It hits the floor 0.45 seconds later. What are the horizontal and vertical components? Vx = 5 m/s t = 0.45 s g = -9.8 m/s2 Vy = ? Rv= ?

  12. Vx= 5 m/s V

  13. PROJECTILES LAUNCHED AT AN ANGLE vi vyi θ vxi

  14. VERTICAL AND HORIZONTAL PROJECTILE MOTION Vy θ Vx V

  15. The path of a projectile is a parabola. θ

  16. The horizontal velocity vx remains constant the whole flight. The vertical initial velocity slows to zero at the maximum height where it reverses direction at the top. Velocity increases as the object returns to the ground. Velocity is positive on the way up and negative on the way down. vy = vi sinθ Vx =vicosθ vx = vi cos θ vy = 0 vx =vi cos θ R vy = -vi sin θ

  17. The angle that a projectile is launched will determine the range and maximum height it obtains. Vy = Visin θ Vx = Vicosθ

  18. Maximum range is an angle of 47º

  19. Projectiles Launched at an Angle Equations X- component and Y-component Equations Rv = Vi Vi = initial velocity projectile was launched Viy Viy= Vi (sinθ) θ Vix Vix = Vi (cosθ)

  20. vi Projectiles Launched at an Angle viy vix Horizontal vix = constant dx = vixt Veritcalvyf = viy+ gt

  21. Example: A golf ball is hit with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find: a. Its position and velocity after 8 s Let’s first find the components of the velocity vix= 100 cos 30 = 86.6 m/s viy= 100 sin 30 = 50 m/s Vi Viy Vix dx = vixt = 86.6(8) = 692.8 m dy = viyt + ½ gt2 = 50(8) + ½ (-10)(8)2 = 80.00 m vx = vix= 86.6 m/s vy = viy+ gt = 50 + (-10)(8) = - 30 m/s

  22. b. The time required to reach its maximum height At top fy = 0 vfy= viy+ gt 0 = 50 m/s + -10m/s2t t = 5.0 s c. The horizontal range dx Total time T = 2t = 2(5.0) = 10.0 s x = vixt = 86.6(10.0) = 866 m

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