Introduction to Projectile Motion

1. Introduction toProjectile Motion Mr.Chin-Sung Lin

2. Introduction to Projectile Motion • What is Projectile Motion? • Trajectory of a Projectile • Calculation of Projectile Motion

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

4. What is Projectile Motion?

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

6. 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.

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

9. Trajectory (Path) of a Projectile

10. Trajectory (Path) of a Projectile

11. y v0 x

12. y x

13. y x

14. y x

15. y • Velocity is changing and the motion is accelerated • The horizontal component of velocity (vx) is constant • Acceleration from the vertical component of velocity (vy) • Acceleration due to gravity is constant, and downward • a = - g = - 9.81m/s2 g = 9.81m/s2 x

16. y • The horizontal and vertical motions are independent of each other • Both motions share the same time (t) • The horizontal velocity ....vx = v0 • The horizontal distance .... dx = vx t • The vertical velocity ........ vy = - g t • The vertical distance ........ dy = 1/2 gt2 g = 9.81m/s2 x

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

18. Trajectory (Path) of a Projectile • H velocity is constant vx = v0 • V velocity is changing vy = - g t • H range: dx = v0 t • V distance: dy = 1/2 gt2

19. Introduction to Projectile Motion • What is Projectile Motion? • Trajectory of a Projectile • Calculation of Projectile Motion

20. v0 d g t R Calculation of Projectile Motion • Example: A projectile was fired with initial velocity v0horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile.

21. v0 d g t R Strategies of Solving Projectile Problems • H & V motions can be calculated independently • H & V kinematics equations share the same variable t

22. v0 d g t R Strategies of Solving Projectile Problems H motion: dx = vx t R = v0 t V motion: dy = d = 1/2 gt2 t =sqrt(2d/g) So, R = v0 t = v0 * sqrt(2d/g)

23. Numerical Example of Projectile Motion H motion: dx = vx t R = v0 t = 10 t V motion: dy = d = 1/2 gt2 t =sqrt(2 *19.62/9.81) = 2 s So, R = v0 t = v0 * sqrt(2d/g) = 10 * 2 = 20 m V0 = 10 m/s g = 9.81 m/s2 19.62 m t R

24. V0 = 10 m/s g = 9.81 m/s2 d t 20 m 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.

25. V0 = 10 m/s g = 9.81 m/s2 d t 20 m Exercise 1: Projectile Problem H motion: dx = vx t 20 = v0 t = 10 t t = 2 s V motion: dy = d = 1/2 gt2 = 1/2 (9.81)22 = 19.62 m So, d = 19.62 m

26. V0 g = 9.81 m/s2 19.62 m t 20 m 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 v0 of the projectile.

27. V0 g = 9.81 m/s2 19.62 m t 20 m Exercise 2: Projectile Problem H motion: dx = vx t 20 = v0 t V motion: dy = d = 1/2 gt2 t =sqrt(2 *19.62/9.81) = 2 s So, 20 = v0 t = 2 v0 v0 = 20/2 = 10 m/s

28. Summary of Projectile Motion • What is Projectile Motion? • Trajectory of a Projectile • Calculation of Projectile Motion

29. Projectile Motion with Angles Mr.Chin-Sung Lin

30. 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/s2 20 m/s vy 60o vx

31. 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? v g = 9.81 m/s2 20 m/s vy t 60o vx R

32. 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/s2 20 m/s vy h 60o vx

33. 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/s2 20 m/s vy t 60o vx

34. 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/s2 20 m/s vy 60o vx R

35. 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/s2 20 m/s vy 60o vfx vx vfy vf

36. 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/s2 20 m/s q R