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PROJECTILE MOTION CHAPTER 3. PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile.

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Presentation on theme: "PROJECTILE MOTION CHAPTER 3. PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile."— Presentation transcript:

1 PROJECTILE MOTION CHAPTER 3

2 PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile.

3 If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.

4 The projectile has a constant horizontal velocity and a vertical velocity that changes uniformly under the influence of the acceleration due to gravity.

5 HORIZONTAL PROJECTION If an object is projected horizontally, its motion can best be described by considering its horizontal and vertical motion separately.

6 In the figure we can see that the vertical velocity and position increase with time as those of a free-falling body. Note that the horizontal distance increases linearly with time, indicating a constant horizontal velocity.

7 3.19 A cannonball is projected horizontally with an initial velocity of 120 m/s from the top of a cliff 250 m above a lake. a. In what time will it strike the water at the foot of the cliff? b. What is the x-distance (range) from the foot of the cliff to the point of impact in the lake? v 0 x = 120 m/s y = 250 m v 0y = 0 = 7.14 s x = v x t = 120(7.14) = 857 m UAM UM

8 c. What are the horizontal and vertical components of its final velocity? d. What is the final velocity at the point of impact and its direction? v x = 120 m/s v y = v oy + gt = (7.14) = 70 m/s = 139 m/s = 30.2  below horizontal v (139 m/s, 30.2  ) UM UAM

9 3.20 A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff? v x = 15 m/s x = 47 m v y = 0 = 3.13 s y = ½ gt 2 = ½ (9.8)(3.13) 2 = 48 m UM UAM

10 PROJECTILE MOTION AT AN ANGLE The more general case of projectile motion occurs when the projectile is fired at an angle.

11 Problem Solution Strategy: 1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = m/s 2 2. Resolve the initial velocity v o into its x and y components: v ox = v o cos θ v oy = v o sin θ 3. The horizontal and vertical components of its position at any instant is given by: x = v ox t y = v oy t +½gt 2 4. The horizontal and vertical components of its velocity at any instant are given by: v x = v ox v y = v oy + gt 5. The final position and velocity can then be obtained from their components.

12 3.23 An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30  above the horizontal. Find: a. Its position and velocity after 8 s v o = 100 m/s, 30  t = 8 s g = m/s 2 v ox = 100 cos 30  = 86.6 m/s v oy = 100 sin 30  = 50 m/s x = v ox t = 86.6(8) = m y = v oy t + ½ gt 2 = 50(8) + ½ (-9.8)(8) 2 = 86.4 m v x = v ox = 86.6 m/s v y = v oy + gt = 50 + (-9.8)(8) = m/s UAM UM

13 b. The time required to reach its maximum height At the top of the path: v y = 0 v y = v oy + gt = 5.1 s c. The horizontal distance (range) Total time T = 2t = 2(5.1) = 10.2 s x = v ox t = 86.6(10.2) = m UM UAM

14 3.24 A baseball is thrown with an initial velocity of 120 m/s at an angle of 40  above the horizontal. How far from the throwing point will the baseball attain its original level? v o = 120 m/s, 40  g = m/s 2 v ox = 120 cos 40  = 91.9 m/s v oy = 120 sin 40  = 77.1 m/s At top v y = 0 = 7.9 s x = v ox (2t) = 91.9(2)(7.9) = 1452 m UAM UM

15 3.25 A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s. a. At what angle with respect to the horizontal was it released? v o = 15 m/s t = 2 s time to maximum height = 1 s at the top v y = 0 v y = v oy + gt = 40.8º UAM

16 b. What was the maximum height achieved by the ball? y = v oy t +½ gt 2 = (15)(sin 40.8º)(1) + ½ (-9.8)(1) 2 = 4.9 m UAM

17 3.26 Find the range of a gun which fires a shell with muzzle velocity v o at an angle θ. At top v y = 0 v y = v oy + gt = v o sin θ - gt Total time = 2t x = v x t K2D

18 sin θ cos θ = ½ sin 2θ b. What is the angle at which the maximum range is possible? Maximum range is 45  since 2θ = 90 

19 sin θ cos θ = ½ sin 2θ b. What is the angle at which the maximum range is possible? Maximum range is 45  since 2θ = 90 

20 c. Find the angle of elevation  of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant. v o = 120 m/s x = 1300 m = sin -1 (2θ)= 62  θ = 31 

21 PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile.

22 If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.

23 c. Find the angle of elevation  of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant. v o = 120 m/s x = 1300 m = sin -1 (2θ)= 62  θ = 31 

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28 c. Find the angle of elevation  of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant. v o = 120 m/s x = 1300 m = sin -1 (2θ)= 62  θ = 31 

29 PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile.

30 If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.

31 c. Find the angle of elevation  of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant. v o = 120 m/s x = 1300 m = sin -1 (2θ)= 62  θ = 31 

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