Projectile Motion

1. Projectile Motion • Concerns the flight of an object or body after it is free of support. • The flight path of a projectile is called the trajectory.

2. Projectile Motion • Concerns the flight of an object or body after it is free of support. • Objects that are continuously being propelled (such as airplanes) aren’t considered projectiles. • http://www.youtube.com/watch?v=3X1ewxVwhug

3. Examples of Projectiles • Football • Javelin • Discus

4. Examples of Projectiles • Long jumper • Diver • High jumper • Rory McIlroy golf shot

5. A human in flight behaves the same projectile laws as any other object.

6. Factors Affecting the Trajectory of a Projectile • The relative height of projection • Projection angle (the initial angle relative to horizontal) • Projection speed (velocity when object first released) Air resistance and wind

7. Factors Affecting the Trajectory of a Projectile Projection speed Projection angle Projection height

8. Relative Projection Height • This is the release height compared to the final landing height of the projectile • Ex. • Relative projection height = 0

9. 2 m Relative Projection Height • Relative projection height = 2 m

10. 10 ft 5.5 ft Relative Projection Height • Relative projection height = -4.5 ft

11. Optimum Angle of Projection (assuming there is no air resistance) • If Relative Projection Height = 0, the optimum angle = 450 • If Relative Projection Height > 0 , the optimum angle < 450 • If Relative Projection Height < 0 , the optimum angle > 450

12. Optimum Projectile Angle • If Relative Projection Height = 0, the optimum angle = 450 • If Relative Projection Height > 0 , the optimum angle < 450 • If Relative Projection Height < 0 , the optimum angle > 450

13. Optimum Angle of Projection:Range

14. Optimum Projection Angles Shot put = 37° Basketball shot = 49-55° Long jump = 20-26° (high forward velocity reduces optimal angle)

15. Components of Projectile Speed By resolving the vertical and horizontal components for the instantaneous velocity vectors, we can find the instantaneous vertical and horizontal velocities.

16. The Components of Projectile Speed EXAMPLE: A ball is thrown upward with a speed of projection of 20 m/s. If the angle of projection is 400, calculate the horizontal and vertical components of the speed of projection. vt = 20 m/s direction: 40° above horiz. vx = Horizontal component vx = (vt)(cos 400) = (20 m/s)(cos 400) vx = 15.32 m/s vy = Vertical component vy = (vt)(sin 400) = (20 m/s)(sin 400) vy = 12.86 m/s vt vy 400 vx

17. Perpendicular Vectors Don’t Directly Affect Each Other • For example, if the projection angle of a projectile is horizontal (the vertical component of the projection speed is 0), it will fall as quickly as if it is dropped with a projection speed of 0.

18. Perpendicular Vectors Don’t Directly Affect Each Other Because the pull of gravity is unaffected by horizontal velocity, a projectile thrown horizontally has the same vertical velocity as an object dropped straight down. If objects are released from the same height they will hit the ground at the same time (neglecting air resistance).

19. Acceleration Due to Gravity in Projectile Motion g (or ag) has the value of 9.81 m/s2 (metric units) or 32 ft/s2 (English units) when used in projectile motion calculations. Because the horizontal and vertical components of a trajectory dont affect each other, if air resistance is neglected horizontal acceleration = 0 and vertical acceleration = g (or ag).

20. Equations of Constant Acceleration • Formulas applied when acceleration is unchanging • v2 = v1 + at [y direction] • d = v1t + (1/2)at2 [y direction] • v22 = v12 + 2a•d • yh = yi + x tanq – agx2/2(vicosq)2 [height] • R = 2 vi2/ag • sin q • cos q These formulas assume that: d= displacement, v1 = initial velocity, R = range v2 = final velocity, a = acceleration, and t = time

21. Laws of Constant Acceleration • R = 2 vi2/ag • sin q • cos q Dave kicks a soccer ball at a velocity of 20 m/sec. If the projectile angle is 30°, find the range. Assume: a) air drag is not significant and b) a relative projectile height = 0 Given: vb = 20 m/sec q = 30° Find: R

22. Equations of Constant Acceleration • Formula: R = 2 vi2/ag • sin q • cos q Solution: R = 2• (20 m/s)2/9.81 m/s2 • sin 30° • cos30° R= 2 • (400m2/s2)/9.81 m/s2 • 0.5 • 0.866 R = 35.31 m 30° Range

23. Air Resistance and Projectile Motion Air resistance will tend to cause a projectile to fall shorter than it would if there were no air resistance. Without Air Resistance With Air Resistance

24. Air Resistance and Projectile Motion Air drag or air resistance creates a vector that is opposite in direction with the net velocity vector of the object at any point along the trajectory. Without Air Resistance Air Drag

25. Air Resistance and Projectile Motion 1) reduces height at apex 2) reduces range (horizontal distance) 3) angle of descent steeper (when compare to projectile angle) **4) If a frisbee or discus are thrown at an orientation of their long axis slightly higher (angle of attack) angle than the projectile angle, the air drag will actually help keep them aloft longer and farther (range) over the second half of the flight Long axis angle of attack Projection angle q

26. Air Resistance and Projectile Motion **4) If a frisbee or discus are thrown at an orientation of their long axis slightly higher (angle of attack) angle than the projectile angle, the air drag will actually help keep them aloft longer and farther (range) over the second half of the flight Long axis angle of attack Projection angle q

27. Air Resistance and Projectile Motion **4) If a frisbee or discus are thrown at an orientation of their long axis slightly higher (angle of attack) angle than the projectile angle, the air drag will actually help keep them aloft longer and farther (range) over the second half of the flight Long axis angle of attack Projection angle q