Ch#4 MOTION IN 2 DIMENSIONS

this is for students

1. Motion in Two Dimensions (Projectile motion) Chapter04 from HRK. Course code: 301 Course Title: Mechanics and properties of Matter Course Incharge : SEHRISH INAM Date: March08,2022

2. Projectile Motion What is the path of a projectile as it moves through the air? Parabolic? Straight up and down? Yes, both are possible . What forces act on projectiles? Only gravity, which acts only in the negative y-direction. Air resistance is ignored in projectile motion.

3. Motion in Two Dimensions Using + or – signs is not always sufficient to fully describe motion in more than one dimension Vectors can be used to more fully describe motion Still interested in displacement, velocity, and acceleration Will serve as the basis of multiple types of motion in future chapters

4. General Motion Ideas In two- or three-dimensional kinematics, everything is the same as as in one-dimensional motion except that we must now use full vector notation Positive and negative signs are no longer sufficient to determine the direction

5. Projectile Motion An object may move in both the x and y directions simultaneously The form of two-dimensional motion we will deal with is called projectile motion The object which is being moved in 2 dimension simultaneously is known as projectile.

6. Assumptions of Projectile Motion The free-fall acceleration g is constant over the range of motion And is directed downward The effect of air friction is negligible With these assumptions, an object in projectile motion will follow a parabolic path This path is called the trajectory

7. Projectile Motion Illustrated 7

8. Verifying the Parabolic Trajectory Reference frame chosen y is vertical with upward positive Acceleration components a y = -g and a x = 0 Initial velocity components v xi = v i cosФ and v yi = v i sin Ф

9. Verifying the Parabolic Trajectory, cont…. Displacements x f = v xi t = ( v i cos Ф ) t y f = v yi t + 1/2 a y t 2 = ( v i sin Ф ) t - 1/2 gt 2 Combining the equations gives : This is in the form of y = ax – bx 2 which is the standard form of a parabola

10. Analyzing Projectile Motion Consider the motion as the superposition of the motions in the x - and y -directions The x -direction has constant velocity a x = 0 The y-direction is free fall a y = - g The actual position at any time is given by: r f = r i + v i t + 1/2 g t 2

11. Projectile Motion Vectors r f = r i + v i t + 1/2 g t 2 The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration

12. Projectile Motion Diagram

13. Projectile Motion Implications The y -component of the velocity is zero at the maximum height of the trajectory The acceleration stays the same throughout the trajectory --------- ------------------

14. Range &Maximum Height of a Projectile When analyzing projectile motion, two characteristics are of special interest The range, R , is the horizontal distance of the projectile The maximum height the projectile reaches is h .

15. Height of a Projectile, equation The maximum height of the projectile can be found in terms of the initial velocity. This equation is valid only for symmetric motion

16. Range of a Projectile, equation The range of a projectile can be expressed in terms of the initial velocity vector: This is valid only for symmetric trajectory

17. More About the Range of a Projectile

18. Range of a Projectile, final The maximum range occurs at Ф = 45 o Complementary angles will produce the same range The maximum height will be different for the two angles The times of the flight will be different for the two angles.

19. Projectile Motion – Problem Solving Hints Select a coordinate system Resolve the initial velocity into x and y components Analyze the horizontal motion using constant velocity techniques Analyze the vertical motion using constant acceleration techniques Remember that both directions share the same time.

20. Non-Symmetric Projectile Motion Follow the general rules for projectile motion Break the y -direction into parts up and down or symmetrical back to initial height and then the rest of the height May be non-symmetric in other ways

21. Acceleration in Different Frames of Reference The derivative of the velocity equation will give the acceleration equation The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame.

22. Projectile Motion Special case of 2-D motion Horizontal motion: a x = 0 so v x = constant Vertical motion: a y = g = constant so the constant acceleration equations apply. Assumptions: Horizontal and vertical motions are independent of each other Air resistance (i.e., drag) can be ignored. 22

23. Projectile Motion Illustrated 23

24. Motion with Constant Acceleration v = v o + at x − x o = v o t + ½ at 2 v 2 = v o 2 + 2 a ( x − x o ) x − x o = ½ ( v o + v ) t x − x o = vt − ½ at 2 24

25. Free-Fall Acceleration Equations If +y is vertically up, then the free-fall acceleration due to gravity near Earth’s surface is a = − g = − 9.8 m/s 2 . v = v o − gt y − y o = v o t − ½ gt 2 v 2 = v o 2 − 2 g ( y − y o ) y − y o = ½ ( v o + v ) t y − y o = vt + ½ gt 2 25

26. The Trajectory of a Projectile What does the free-body diagram look like for force? F g

27. The Vectors of Projectile Motion What vectors exist in projectile motion? Velocity in both the x and y directions. Acceleration in the y direction only. v y (Increasing) v x (constant) a y = -9.8m/s 2 a x = 0 Trajectory or Path Why is the velocity constant in the x-direction? No force acting on it. Why does the velocity increase in the y-direction? Gravity.

28. Ex. 1: Launching a Projectile Horizontally A cannonball is shot horizontally off a cliff with an initial velocity of 30 m/s. If the height of the cliff is 50 m: How far from the base of the cliff does the cannonball hit the ground? With what speed does the cannonball hit the ground?

29. Diagram the problem 50m F g = F net a = -g v i = 30m/s v f = ? v x v y x = ?

30. State the Known & Unknown Known: v ix = 30 m/s v iy = 0 m/s a = -g = -9.81m/s 2 d y = -50 m Unknown: d x at y = -50 m v f = ?

31. Perform Calculations (y) Strategy: Use reference table to find formulas you can use. v fy = v iy + gt d y = v iy t + ½ gt 2 Note that g has been substituted for a and y for d. Use known factors such as in this case where the initial velocity in the y-direction is known to be zero to simplify the formulas. v fy = v iy + gt v fy = gt ( 1) d y = v iy t + ½ gt 2 d y = ½ gt 2 (2 )

32. Perform Calculations (y) Cont… (Use the second formula (2) first because only time is unknown)

33. Perform Calculations (y ) Cont….. Now that we have time, we can use the first formula (1) to find the final velocity. v fy = gt v y = (-9.8 m/s 2 )(3.2 s) = -31 m/s

34. Perform Calculations (x) Strategy: Since you know the time for the vertical(y-direction), you also have it for the x-direction. Time is the only variable that can transition between motion in both the x and y directions. Since we ignore air resistance and gravity does not act in the horizontal (x-direction), a = 0. Choose a formula from your reference table d x = v ix t + ½ at 2 Since a = 0, the formula reduces to x = v ix t d x = (30 m/s)(3.2 s) = 96 m from the base.

35. Finding the Final Velocity ( v f ) We were given the initial x-component of velocity, and we calculated the y-component at the moment of impact. Logic: Since there is no acceleration in the horizontal direction, then v ix = v fx . We will use the Pythagorean Theorem. v fx = 30m/s v fy = -31m/s v f = ?

36. Ex. 2: Projectile Motion above the Horizontal A ball is thrown from the top of the Science Wing with a velocity of 15 m/s at an angle of 50 degrees above the horizontal. What are the x and y components of the initial velocity? What is the ball’s maximum height? If the height of the Science wing is 12 m, where will the ball land?

37. Diagram the problem y x v i = 15 m/s  = 50 ° a y = -g Ground 12 m x = ? v i = 15 m/s v ix v iy  = 50 °

38. State the Known & Unknown Known: d yi = 12 m v i = 15 m/s  = 50° a y = g = -9.8m/s 2 Unknown: d y (max) = ? t = ? d x = ? v iy = ? v ix = ?

39. Perform the Calculations ( y max ) y-direction: Initial velocity: v iy = v i sin  v iy = (15 m/s)(sin 50 °) v iy = 11.5 m/s Time when v fy = 0 m/s: v fy = v iy + gt ( ball at peak ) t = v iy / g t = (-11.5 m/s)/(-9.81 m/s 2 ) t = 1.17 s Determine the maximum height: d y (max) = y i + v iy t + ½ gt 2 d y (max) = 12 m + (11.5 m/s)(1.17 s) + ½ (-9.81 m/s 2 )(1.17 s) 2 d y (max) = 18.7 m v i = 15 m/s v xi v yi  = 50 °

40. Perform the Calculations (t) Since the ball will accelerate due to gravity over the distance it is falling back to the ground, the time for this segment can be determined as follows Time from peak to when ball hits the ground: From reference table: d y (max) = v iy t + ½ gt 2 Since y i can be set to zero as can v iy , t = 2* d y (max) /g t = 2(-18.7 m)/(-9.81 m/s 2 ) t = 1.95 s By adding the time it takes the ball to reach its maximum height (peak) to the time it takes to reach the ground will give you the total time. t total = 1.17 s + 1.95 s = 3.12 s  

41. Perform the Calculations (x) x-direction: Initial velocity: v ix = v i cos  v ix = (15 m/s)( cos 50 °) v ix = 9.64 m/s Determine the total distance: x = v ix t d x = (9.64 m/s)(3.12 s) d x = 30.1 m v i = 15 m/s v xi v yi  = 50 °

42. Analyzing Motion in the x and y directions independently. x-direction y-direction d x = v ix t = v fx t d y = ½ (v i + v f ) t d y = v avg t v ix = v i  cos  v f = v iy + g t d y = v iy t + ½g(t) 2 v fy 2 = v iy 2 + 2g d y v iy = v i  sin 