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Projectile Motion

This article discusses the resultant velocity of a boat with a constant motor speed and current, and explores the concepts of relative velocity and projectile motion. Includes examples and equations.

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Projectile Motion

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  1. Projectile Motion

  2. First, let’s talk about Mr. Anz’sboat… . If the boat has a constant motor speed of 10 m/s and the current is 5 m/s, what is the resultant velocity of the boat?

  3. If the boat has a constant motor speed of 10.0 m/s and the current is 5.0 m/s, what is the resultant velocity of the boat? C2 = (10m/s)2 + (5m/s)2 C= 11.18 m/s 5 m/s θ = 5 m/s 10 m/s 10 m/s Θ = 26.57° 26.57° C= 11.18 m/s with respect to the bank

  4. RELATIVE VELOCITY Velocity measurements differ in different frames of reference. If the frame of reference is denoted with subscripts (vab is the velocity of a with respect to b), then the velocity of an object with respect to a different frame of reference can be found by adding the known velocities.  vab = vac + vcb

  5. If the boat has a speed of 10.0 m/s and the current is 5.0 m/s, what is the resultant velocity of the boat? How long to travel across the 120.0 m wide river? VECTORS ARE INDEPENDENT The time to cross depends on the speed across the river. t = d v = 120.0 m 10.0m/s = 12 sec How far downstream will the boat land on the far bank? The distance downstream depends on the downstream current speed and the time in the water. d = vt = (5.0 m/s)(12sec) = 60. m downstream

  6. The BIG Rule about vectors; The perpendicular components of motion are INDEPENDENT of each other So… the velocity across the river is independent of the velocity down the river. We will use this rule again and again…

  7. A PROJECTILE is an object that moves through space acted upon only by the earth’s gravity. Dropped Throw up Launched at an angle…

  8. facts about projectiles; 1. Projectiles maintain a constant horizontal velocity (neglecting air resistance) 2. Projectiles always experience a constant vertical accelerationof “g” or 9.8 m/s2 (neglecting air resistance)

  9. 3. Horizontal and vertical motion are completely INDEPENDENT of each other. So we can separate the velocity of a projectile into horizontal and vertical components.

  10. What about equations? All of the kinematic equations that you used in one dimension still work! Accelerated motion: vf = vo + at vf2 = v02 + 2ad d = vot + ½ at2 Constant Velocity: d = vt

  11. Let’s apply these rules to a horizontal thrown projectile… A ball is thrown horizontally from a 100 m building at a velocity of 20 m/s. How far from the base of the building will it land? For vertical; dy = 100 m V0 = 0 m/s g = 9.8 m/s2 For horizontal; dx = ? Vx = 20 m/s a = 0 We can use the vertical to find time… 100m Since horizontal distance … dx = Vxt dy = Vot + ½at2 (4.52s) dx = (20m/s) ? 100m = 0 + ½(9.8m/s2)t2 dx = 90.35 m t = 4.52 s

  12. Example 3 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 = 15.0 m/s x = 47 m vyo = 0 m/s = 3.13 s dy = vot + ½ gt2 = ½ (9.8m/s2)(3.13s)2 = 48.11 m

  13. Let’s look at an example of an up/down problem. V = 0 m/s - 9.8 m/s2 + 9.8 m/s2 up down Rule 4: If the start height and the end height are the same, then time up equals the time down and the distance up equals the distance down

  14. A plane flying at 115 m/s drops a package from 600m. How far from the drop point will it land? Rule 4. Objects dropped from a moving vehicle have the same horizontal velocityas the moving vehicle.

  15. Horizontal: Vx = 115 m/s dx = ? Vertical: Voy = 0 dy = 600. m a = 9.8 m/s2 This is the same problem we’ve been working… dy = ½ at2 dx = (115 m/s)(11.07s) 600. m= ½ (9.8m/s2)t2 dx = 1272.55 m t = 11.07 s

  16. One more… A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. Horizontal Vertical d = Vot + ½at2 Vx = ? dx = 35.0m Voy = 0 dy = 22.0m a = 9.8 m/s2 22.0 m = 0 + ½(9.8 m/s2)t2 t = 2.12 s = 35.0 m 2.12s Vx = d t Vx = 16.52 m/s

  17. What is the vertical velocity just at impact? (Vyf) Horizontal Vertical Voy = 0 dy = 22.0m a = 9.8 m/s2 Vfy = ? Vx = 16.52m/s dx = 35.0m t = 2.12 s Vfy = Vo + at Vfy = 0 + (9.8m/s2)(2.12s) Vfy = 20.77 m/s What is the resultant velocity of the ball at impact? VR2 = (16.5m/s)2+ (20.77 m/s)2 = 26.53 m/s VR Vfy Tan θ = 20.77 m/s 16.52 m/s = 51.5º to the ground θ Vx

  18. facts about projectiles; 1. Projectiles maintain a constant horizontal velocity (neglecting air resistance) 2. Projectiles always experience a constant vertical acceleration of “g” or 9.8m/s2 (neglecting air resistance) 3. Horizontal and vertical motion are completely INDEPENDENT of each other. 4. If the start height and the end height are the same, then time up equals the time down and the distance up equals the distance down 5. Objects dropped from a moving vehicle have the same horizontal velocityas the moving vehicle.

  19. This evening… you will practice using these rules of projectile motion. Wednesday we will add in projectiles that are launched at an angle.. • Your homework: • Problems 1-7

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

  21. Problem solution: 1. Upward direction is positive. Acceleration (g) is downward thus negative. 2. Resolve the initial velocity vo into its x and y components: and 3. The horizontal and vertical components of its position at any instant is given by: and 4. The horizontal and vertical components of its velocity at any instant are given by: and 5. The finalposition and velocity can then be obtained from their components.

  22. Example 6A ball is thrown from the top of one building toward a tall building 50.0 m away. The initial velocity of the ball is 20. 0m/s at 40. How far above or below its original level will the ball strike the opposite wall? x = 50.0 m vo = 20.0 m/s, 40 g = - 9.8 m/s2 vox = 20.0 m/s cos 40 = 15.3 m/s voy = 20.0 sin 40 = 12.9 m/s = 3.26 s y = voy t + ½ gt2 = (12.9 m/s)(3.26s) + ½ (-9.8m/s2)(3.26s)2 = - 10.2 m or 10.2 m below its original level

  23. Example 7An 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 8.0 s vo = 100 m/s, 30 t = 8.0 s g = - 9.8 m/s2 vox = 100. m/s cos 30 = 86.6 m/s voy = 100. sin 30 = 50.0 m/s vx = vox = 86.6 m/s vy = voy + gt = 50.0 m/s + (-9.8m/s2)(8.0s) = - 28.4 m/s x = vox t = (86.6 m/s)(8.0s) = 692.82 m y = voy t + ½ gt2 = (50.0 m/s)(8.0s) + ½ (-9.8m/s2)(8.0s)2 = 86.4 m

  24. b. The time required to reach its maximum height At top vy = 0vy = voy + gt = 5.10 s c. The horizontal range R Total time T = 2t = 2(5.10s) = 10.2 s x = vox t = 86.6 m/s (10.2 s) = 883 m

  25. Example 8A baseball is thrown with an initial velocity of 120. m/s at an angle of 40.0above the horizontal. How far from the throwing point will the baseball attain its original level? vox = 120. cos 40.0 = 91.9 m/s voy = 120. m/s sin 40.0 = 77.1 m/s vo = 120. m/s, 40.0 g = - 9.8 m/s2 At top vy = 0 = 7.87 s x = vox (2t) = (91.9 m/s)(2)(7.87s) = 1447.06 m

  26. Example 9.Find the range of a gun which fires a shell with muzzle velocity v at an angle . What is the maximum range possible? At top vy = 0 vy = voy + gt = vo sin θ - gt Total time = 2t x = vxt

  27. sin θcos θ= ½ sin 2θ Maximum range is 45 since 2θ = 90

  28. b. 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

  29. FastCheck! Pencil and Calculator, please

  30. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity A) decreases. B) remains constant. C) increases.

  31. As a bowling ball rolls off a table the horizontal velocity remains constant. There is no acceleration in the horizontal direction.

  32. A bullet fired horizontally from a rifle begins to fall A) as soon as it leaves the barrel. B) after air friction reduces its speed. C) neither of these

  33. A bullet fired horizontally begins to fall as soon as the bullet leaves the bullet leaves the barrel. Gravity pulls on the bullet immediately.

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