2. Physics Lesson 6 Projectile Motion Eleanor Roosevelt High School Mr. Chin-Sung Lin

3. Introduction to Projectile Motion  What is Projectile Motion?  Trajectory of a Projectile  Calculation of Projectile Motion

4. Introduction to Projectile Motion  What is Projectile Motion?  Trajectory of a Projectile  Calculation of Projectile Motion

5. What is Projectile Motion?

6. Features of Projectile Motion? 2-D Motion Parabolic Path Affected by Gravity Thrown into the Air Determined by Initial Velocity

7. Definition: Projectile Motion Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

8. Definition: Projectile Motion Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

9. Introduction to Projectile Motion  What is Projectile Motion?  Trajectory of a Projectile  Calculation of Projectile Motion

10. Trajectory (Path) of a Projectile

12. v0v0 x y

13. x y

14. x y

15. x y

16. x y  Velocity is changing and the motion is accelerated  The horizontal component of velocity (v x ) is constant  Acceleration from the vertical component of velocity (v y )  Acceleration due to gravity is constant, and downward  a = g = - 9.81m/s 2 g = -9.81m/s 2

18. x y  The horizontal and vertical motions are independent of each other  Both motions share the same time (t)  The horizontal velocity....v x = v 0  The horizontal distance.... d x = v x t  The vertical velocity........ v y = g t  The vertical distance........ d y = 1 / 2 g t 2 g = -9.81m/s 2

19. Trajectory (Path) of a Projectile  T he path of a projectile is the result of the simultaneous effect of the H & V components of its motion  V component  accelerated downward motion  H component  constant velocity motion & V motions are independent & V motions share the same time t  T he projectile flight time t is determined by the V component of its motion

20. Horizontally Launched Projectile  H velocity is constant v x = v 0  V velocity is changing v y = g t  H range: d x = v 0 t  V distance: d y = 1 / 2 g t 2

21. Introduction to Projectile Motion  What is Projectile Motion?  Trajectory of a Projectile  Calculation of Projectile Motion

22. Calculation of Projectile Motion  Example: A projectile was fired with initial velocity v 0 horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile. g R d v0v0 t

23. Strategies of Solving Projectile Problems  H & V motions can be calculated independently  H & V kinematics equations share the same variable t g R d v0v0 t

24. Strategies of Solving Projectile Problems H motion: d x = v x t R = v 0 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2d/g) So, R = v 0 t = v 0 * sqrt(2d/g) g R d v0v0 t

25. Numerical Example of Projectile Motion H motion: d x = v x t R = v 0 t = 10 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2 *19.62/9.81) = 2 s So, R = v 0 t = v 0 * sqrt(2d/g) = 10 * 2 = 20 m g = 9.81 m/s 2 R 19.62 m V 0 = 10 m/s t

26. Exercise 1: Projectile Problem A projectile was fired with initial velocity 10 m/s horizontally from a cliff. If the horizontal range of the projectile is 20 m, calculate the height d of the cliff. g = 9.81 m/s 2 20 m d V 0 = 10 m/s t

27. Exercise 1: Projectile Problem H motion: d x = v x t 20 = v 0 t = 10 t t = 2 s V motion: d y = d = 1 / 2 g t 2 = 1 / 2 (9.81) 2 2 = 19.62 m So, d = 19.62 m g = 9.81 m/s 2 20 m d V 0 = 10 m/s t

28. Exercise 2: Projectile Problem A projectile was fired horizontally from a cliff 19.62 m above the ground. If the horizontal range of the projectile is 20 m, calculate the initial velocity v 0 of the projectile. g = 9.81 m/s 2 20 m 19.62 m V0V0 t

29. Exercise 2: Projectile Problem H motion: d x = v x t 20 = v 0 t V motion: d y = d = 1 / 2 g t 2 t = sqrt(2 *19.62/9.81) = 2 s So, 20 = v 0 t = 2 v 0 v 0 = 20/2 = 10 m/s g = 9.81 m/s 2 20 m 19.62 m V0V0 t

30. Summary of Projectile Motion  What is Projectile Motion?  Trajectory of a Projectile  Calculation of Projectile Motion

31. Projectile Motion with Angles

32. Example: Projectile Problem – H & V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the horizontal and vertical components of the initial velocity? g = 9.81 m/s 2 20 m/s 60 o vxvx vyvy

33. Example: Projectile Problem – H & V A projectile was fired from ground with 20. m/s initial velocity at 60-degree angle. What’s the horizontal and vertical components of the initial velocity? g = 9.81 m/s 2 20 m/s 60 o vxvx vyvy V x = V cos θ = 20 m/s cos 60 o = 10 m/s V y = V sin θ = 20 m/s sin 60 o = 17.32 m/s

34. Example: Projectile Problem – At the Top A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the velocity of the projectile at the top of its trajectory? g = 9.81 m/s 2 R v t 20 m/s 60 o vyvy vxvx

35. Example: Projectile Problem – At the Top A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the velocity of the projectile at the top of its trajectory? g = 9.81 m/s 2 R v t 20 m/s 60 o vyvy vxvx V = V x = 10 m/s

36. Example: Projectile Problem – Height A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the maximum height that the ball can reach? g = 9.81 m/s 2 h 20 m/s 60 o vyvy vxvx

37. Example: Projectile Problem – Height A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the maximum height that the ball can reach? g = 9.81 m/s 2 h 20 m/s 60 o vyvy vxvx V f 2 = V i 2 + 2gd (0 m/s) 2 = (17.32 m/s) 2 + 2 (-9.81 m/s 2 ) d d = 15.29 m

38. Example: Projectile Problem - Time A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How long will the ball travel before hitting the ground? g = 9.81 m/s 2 t 20 m/s 60 o vyvy vxvx

39. Example: Projectile Problem - Time A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How long will the ball travel before hitting the ground? g = 9.81 m/s 2 t 20 m/s 60 o vyvy vxvx V f = V i + gt 0 m/s = 17.32 m/s + (-9.81 m/s 2 ) t t = 1.77 s 1.77 s x 2 = 3.53 s

40. Example: Projectile Problem – H Range A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How far will the ball reach horizontally? g = 9.81 m/s 2 R 20 m/s 60 o vyvy vxvx

41. Example: Projectile Problem – H Range A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How far will the ball reach horizontally? g = 9.81 m/s 2 R 20 m/s 60 o vyvy vxvx d x = V x t (R = 10 m/s )(3.53 s) = 35.3 m

42. Example: Projectile Problem – Final V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the final velocity of the projectile right before hitting the ground? g = 9.81 m/s 2 20 m/s 60 o vyvy vxvx v fx v fy vfvf

43. Example: Projectile Problem – Final V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the final velocity of the projectile right before hitting the ground? g = 9.81 m/s 2 20 m/s 60 o vyvy vxvx v fx v fy vfvf V fx = V x V fx = 10 m/sV fy = -V y V fx = -17.32 m/s V f = sqrt (V fx 2 + V fy 2 ) = 20 m/s θ = tan -1 (V fy /V fx ) = -60 o

44. Example: Projectile Problem – Max R A projectile was fired from ground with 20 m/s initial velocity. How can the projectile reach the maximum horizontal range? What’s the maximum horizontal range it can reach? g = 9.81 m/s 2 R 20 m/s 

45. V & H Velocity Vectors of Projectile

46. 15 o 30 o 45 o 60 o 75 o Launch Angles of Projectile

47. The End