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Projectile Motion Chapter 3

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

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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.

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

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

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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.

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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? UAM v0x = 120 m/s y = 250 m v0y = 0 = 7.14 s b. What is the x-distance (range) from the foot of the cliff to the point of impact in the lake? UM x = vx t = 120(7.14) = 857 m

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**vx = 120 m/s vy = voy + gt = 0 + 9.8 (7.14) = 70 m/s = 139 m/s**

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

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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? vx = 15 m/s x = 47 m vy = 0 UM = 3.13 s y = ½ gt2 = ½ (9.8)(3.13)2 = 48 m UAM

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**PROJECTILE MOTION AT AN ANGLE**

The more general case of projectile motion occurs when the projectile is fired at an angle.

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**Problem Solution Strategy:**

1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = m/s2 2. Resolve the initial velocity vo into its x and y components: vox = vo cos θ voy = vo sin θ 3. The horizontal and vertical components of its position at any instant is given by: x = voxt y = voy t +½gt2 4. The horizontal and vertical components of its velocity at any instant are given by: vx = vox vy = voy + gt 5. The final position and velocity can then be obtained from their components.

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**vox = 100 cos 30 = 86.6 m/s voy = 100 sin 30 = 50 m/s x = vox t**

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 UAM UM vox = 100 cos 30 = 86.6 m/s voy = 100 sin 30 = 50 m/s vo = 100 m/s, 30 t = 8 s g = m/s2 x = vox t = 86.6(8) = m y = voy t + ½ gt2 = 50(8) + ½ (-9.8)(8)2 = 86.4 m vx = vox = 86.6 m/s vy = voy + gt = 50 + (-9.8)(8) = m/s

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**At the top of the path: vy = 0 vy = voy + gt = 5.1 s Total time T = 2t**

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

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**vox = 120 cos 40 = 91.9 m/s voy = 120 sin 40 = 77.1 m/s**

3.24 A baseball is thrown with an initial velocity of 120 m/s at an angle of 40above the horizontal. How far from the throwing point will the baseball attain its original level? vox = 120 cos 40 = 91.9 m/s voy = 120 sin 40 = 77.1 m/s vo = 120 m/s, 40 g = m/s2 At top vy = 0 UM UAM = 7.9 s x = vox (2t) = 91.9(2)(7.9) = 1452 m

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**time to maximum height = 1 s at the top vy = 0 vy = voy + gt**

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? vo = 15 m/s t = 2 s time to maximum height = 1 s at the top vy = 0 vy = voy + gt UAM = 40.8º

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**y = voy t +½ gt2 = (15)(sin 40.8º)(1) + ½ (-9.8)(1)2 = 4.9 m**

b. What was the maximum height achieved by the ball? UAM y = voy t +½ gt2 = (15)(sin 40.8º)(1) + ½ (-9.8)(1)2 = 4.9 m

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**At top vy = 0 vy = voy + gt x = vxt = vo sin θ - gt Total time = 2t**

3.26 Find the range of a gun which fires a shell with muzzle velocity vo at an angle θ . K2D At top vy = 0 vy = voy + gt = vo sin θ - gt x = vxt Total time = 2t

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**Maximum range is 45 since 2θ = 90**

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

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**Maximum range is 45 since 2θ = 90**

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

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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. vo = 120 m/s x = 1300 m = 0.885 sin-1(2θ)= 62 θ = 31

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

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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.

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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. vo = 120 m/s x = 1300 m = 0.885 sin-1(2θ)= 62 θ = 31

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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. vo = 120 m/s x = 1300 m = 0.885 sin-1(2θ)= 62 θ = 31

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

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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.

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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. vo = 120 m/s x = 1300 m = 0.885 sin-1(2θ)= 62 θ = 31

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