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Chapter 6B – Projectile Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation.

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Presentation on theme: "Chapter 6B – Projectile Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation."— Presentation transcript:

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2 Chapter 6B – Projectile Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

3 Objectives: After completing this module, you should be able to:  Describe the motion of a projectile by treating horizontal and vertical components of its position and velocity. Solve for position, velocity, or time when given initial velocity and launch angle.Solve for position, velocity, or time when given initial velocity and launch angle.

4 Projectile Motion A projectile is a particle moving near the Earth’s surface under the influence of its weight only (directed downward). a = g W WW

5 Vertical and Horizontal Motion Simultaneously dropping yellow ball and projecting red ball horizontally. Click right to observe motion of each ball.

6 Vertical and Horizontal Motion Simultaneously dropping a yellow ball and projecting a red ball horizontally. Why do they strike the ground at the same time? Once motion has begun, the downward weight is the only force on each ball. W W

7 Ball Projected Horizontally and Another Dropped at Same Time: 0 s v ox Vertical Motion is the Same for Each Ball 1 s 2 s 3 s vyvyvyvy vxvxvxvx vxvxvxvx vxvxvxvx vyvyvyvy vyvyvyvy vyvyvyvy vyvyvyvy vyvyvyvy

8 Observe Motion of Each Ball 0 s v ox Vertical Motion is the Same for Each Ball 3 s 2 s 1 s

9 Consider Horizontal and Vertical Motion Separately: Compare Displacements and Velocities 0 s 1 s v ox 2 s 3 s 1 s vyvyvyvy 2 s vxvxvxvx vyvyvyvy 3 s vxvxvxvx vyvyvyvy Horizontal velocity doesn’t change. Vertical velocity just like free fall. vxvxvxvx

10 Displacement Calculations for Horizontal Projection: For any constant acceleration: Horizontal displacement: Vertical displacement: For the special case of horizontal projection:

11 Velocity Calculations for Horizontal Projection (cont.): For any constant acceleration: Horizontal velocity: Vertical velocity: For the special case of a projectile:

12 Example 1: A baseball is hit with a horizontal speed of 25 m/s. What is its position and velocity after 2 s? First find horizontal and vertical displacements: x = 50.0 m y = m 25 m/s x y m +50 m

13 Example 1 Cont.): What are the velocity components after 2 s? 25 m/s Find horizontal and vertical velocity after 2 s: v x = 25.0 m/s v y = m/s vxvx vyvy v 0x = 25 m/s v 0y = 0

14 Consider Projectile at an Angle: A red ball is projected at an angle . At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction). Note vertical and horizontal motions of balls  v oy v ox vovo v x = v ox = constant

15 Displacement Calculations For General Projection: The components of displacement at time t are: For projectiles: Thus, the displacement components x and y for projectiles are:

16 Velocity Calculations For General Projection: The components of velocity at time t are: For projectiles: Thus, the velocity components v x and v y for projectiles are:

17 Problem-Solving Strategy: 1.Resolve initial velocity v o into components: vovo v ox v oy  2. Find components of final position and velocity: Displacement: Velocity:

18 Problem Strategy (Cont.): 3. The final position and velocity can be found from the components. R x y  4. Use correct signs - remember g is negative or positive depending on your initial choice. vovo v ox v oy 

19 Example 2: A ball has an initial velocity of 160 ft/s at an angle of 30 o with horizontal. Find its position and velocity after 2 s and after 4 s. v oy 160 ft/s v ox 30 o Since v x is constant, the horizontal displacements after 2 and 4 seconds are: x = 277 ft x = 554 ft

20 Note: We know ONLY the horizontal location after 2 and 4 s. We don’t know whether it is on its way up or on its way down. x 2 = 277 ft x 4 = 554 ft Example 2: (Continued) v oy 160 ft/s v ox 30 o 277 ft 554 ft 2 s 4 s

21 Example 2 (Cont.): Next we find the vertical components of position after 2 s and after 4 s. v oy = 80 ft/s 160 ft/s  0 s3 s2 s1 s4 s g = -32 ft/s 2 y2y2 y4y4 The vertical displacement as function of time: Observe consistent units.

22 (Cont.) Signs of y will indicate location of displacement (above + or below – origin). v oy = 80 ft/s 160 ft/s  0 s3 s2 s1 s4 s g = -32 ft/s 2 y2y2 y4y4 Vertical position: 96 ft 16 ft Each above origin (+)

23 (Cont.): Next we find horizontal and vertical components of velocity after 2 and 4 s. Since v x is constant, v x = 139 ft/s at all times. Vertical velocity is same as if vertically projected: At any time t: v oy 160 ft/s v ox 30 o

24 v 2y = 16.0 ft/s v 4y = ft/s Example 2: (Continued) v y = 80.0 ft/s 160 ft/s  0 s3 s2 s1 s4 s g = -32 ft/s 2 v2v2 v4v4 At any time t:

25 At 2 s: v 2x = 139 ft/s; v 2y = ft/s Example 2: (Continued) v y = 80.0 ft/s 160 ft/s  0 s3 s2 s1 s4 s g = -32 ft/s 2 v2v2 v4v4 Moving Up +16 ft/s Moving down -48 ft/s The signs of v y indicate whether motion is up (+) or down (-) at any time t. At 4 s: v 4x = 139 ft/s; v 4y = ft/s

26 (Cont.): The displacement R 2,  is found from the x 2 and y 2 component displacements.  0 s2 s 4 s y 2 = 96 ft x 2 = 277 ft R2R2 R 2 = 293 ft   = t = 2 s

27 (Cont.): Similarly, displacement R 4,  is found from the x 4 and y 4 component displacements. R 4 = 558 ft   =  0 s 4 s y 4 = 64 ft x 4 = 554 ft R4R4 t = 4 s

28 (Cont.): Now we find the velocity after 2 s from the components v x and v y. v 2 = 140 ft/s   = v oy = 80.0 ft/s 160 ft/s  0 s2 s g = -32 ft/s 2 v2v2 Moving Up +16 ft/s v 2x = 139 ft/s v 2y = ft/s

29 (Cont.) Next, we find the velocity after 4 s from the components v 4x and v 4y. v 4 = 146 ft/s   = v oy = 80.0 ft/s 160 ft/s  0 s4 s g = -32 ft/s 2 v4v4 v 4x = 139 ft/s v 4y = ft/s

30 Example 3: What are maximum height and range of a projectile if v o = 28 m/s at 30 0 ? y max occurs when 14 – 9.8t = 0 or t = 1.43 s Maximum y-coordinate occurs when v y = 0: v oy 28 m/s v ox 30 o y max v y = 0 v ox = 24.2 m/s v oy = + 14 m/s

31 Example 3(Cont.): What is maximum height of the projectile if v = 28 m/s at 30 0 ? Maximum y-coordinate occurs when t = 1.43 s: y max = 10.0 m v oy 28 m/s v ox 30 o y max v y = 0 v ox = 24.2 m/s v oy = + 14 m/s

32 Example 3(Cont.): Next, we find the range of the projectile if v = 28 m/s at The range x r is defined as horizontal distance coinciding with the time for vertical return. v oy 28 m/s v ox 30 o v ox = 24.2 m/s v oy = + 14 m/s Range x r The time of flight is found by setting y = 0: (continued)

33 Example 3(Cont.): First we find the time of flight t r, then the range x r. v oy 28 m/s v ox 30 o v ox = 24.2 m/s v oy = + 14 m/s Range x r (Divide by t) x r = v ox t = (24.2 m/s)(2.86 s); x r = 69.2 m

34 Example 4: A ball rolls off the top of a table 1.2 m high and lands on the floor at a horizontal distance of 2 m. What was the velocity as it left the table? 1.2 m 2 m First find t from y equation: 0 ½(-9.8)t 2 = -(1.2) t = s Note: x = v ox t = 2 m y = v oy t + ½a y t 2 = -1.2 m R

35 Example 4 (Cont.): We now use horizontal equation to find v ox leaving the table top. Use t = s in x equation: v = 4.04 m/s 1.2 m 2 m R Note: x = v ox t = 2 m y = ½gt 2 = -1.2 m The ball leaves the table with a speed:

36 Example 4 (Cont.): What will be its speed when it strikes the floor? v y = 0 + (-9.8 m/s 2 )(0.495 s) v y = v y + gt 0 v x = v ox = 4.04 m/s Note: t = s v y = m/s v 4 = 146 ft/s   = m 2 m vxvx vyvy

37 Example 5. Find the “hang time” for the football whose initial velocity is 25 m/s, v o =25 m/s 60 0 y = 0; a = -9.8 m/s 2 Time of flight t v ox = v o cos  v oy = v o sin  Initial v o : V ox = (25 m/s) cos 60 0 ; v ox = 12.5 m/s V oy = (25 m/s) sin 60 0 ; v ox = 21.7 m/s Only vertical parameters affect hang time.

38 v o =25 m/s 60 0 y = 0; a = -9.8 m/s 2 Time of flight t v ox = v o cos  v oy = v o sin  Initial v o : 4.9 t 2 = 21.7 t 4.9 t = 21.7 t = 4.42 s Example 5 (Cont.) Find the “hang time” for the football whose initial velocity is 25 m/s, 60 0.

39 Example 6. A running dog leaps with initial velocity of 11 m/s at What is the range? v = 11 m/s  =30 0 Draw figure and find components: v ox = 9.53 m/s v oy = 5.50 m/s v ox = 11 cos 30 0 v oy = 11 sin 30 0 To find range, first find t when y = 0; a = -9.8 m/s t 2 = 5.50 t t = 1.12 s 4.9 t = 5.50

40 Example 6 (Cont.) A dog leaps with initial velocity of 11 m/s at What is the range? v = 10 m/s  =31 0 Range is found from x-component: v x = v ox = 9.53 m/s x = v x t; t = 1.12 s v ox = 10 cos 31 0 v oy = 10 sin 31 0 Horizontal velocity is constant: v x = 9.53 m/s Range: x = 10.7 m x = (9.53 m/s)(1.12 s) = 10.7 m

41 Summary for Projectiles: 1. Determine x and y components v 0 2. The horizontal and vertical components of displacement at any time t are given by:

42 Summary (Continued): 4. Vector displacement or velocity can then be found from the components if desired: 3. The horizontal and vertical components of velocity at any time t are given by:

43 CONCLUSION: Chapter 6B Projectile Motion


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