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Vocab See side board

Rules 4 & 5. Vocab See side board. Rules 1-3. T - Chart. Formulas. Review (River Boat). @ angle #1. Dropped #1. Launched Horizontally #1. Launched Horizontally #2. @ angle #2. Dropped #2. Projectile Motion. ?. Galileo Galilei (1564-1642).

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Vocab See side board

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  1. Rules 4 & 5 Vocab See side board Rules 1-3 T - Chart Formulas Review (River Boat)

  2. @ angle #1 Dropped #1 Launched Horizontally #1 Launched Horizontally #2 @ angle #2 Dropped #2

  3. Projectile Motion ? .

  4. Galileo Galilei (1564-1642) Galileo is alleged to have performed free-fall experiments by dropping objects off the Tower of Pisa

  5. ACCELERATION DUE TO GRAVITY All bodies in free fall near the Earth's surface have the same downward acceleration of: g = 9.8 m/s2 g = 10 m/s2 NOTE: A body in free fall has the same downward acceleration whether it starts from rest or has an initial velocity in some direction.

  6. The presence of air affects the motion of falling bodies. Thus two different objects falling in air from the same height will not, in general, reach the ground at exactly the same time. Because air resistance increases with velocity, eventually a falling body reaches a terminal velocity that depends on its mass, size, shape, and it cannot fall any faster than that.

  7. FREE FALL When air resistance can be neglected, a falling body has the constant acceleration g, and the equations for uniformly accelerated motion apply. Just substitute a for g. Sign Convention for direction of motion: If the object is thrown downwards then g = 10 m/s2 If the object is thrown upwards then g = - 10 m/s2

  8. A ball is dropped from the top of a cliff…What happens to the speed each second? gt Vf = 0 + (10 m/s2)(1s) = 10 m/s Vf = 10m/s + (10 m/s2)(1s) = 20 m/s or Vf = 0 + (10 m/s2)(2s) = 20 m/s

  9. A ball is dropped from the top of a cliff…What happens to the distance traveled each second? d d = 0 + ½ (10 m/s2)(1s)2 = 5 m d = 0 + ½ (10 m/s2)(2s)2 = 20 m

  10. Example 1: A ball is dropped from rest at a height of 50 m above the ground. a. What is its speed just before it hits the ground? vo = 0 m/s d = 50 m g = 10 m/s2 d (0 m/s)2 + 2(10 m/s2)(50m) Vf = 31.62 m/s b. How long does it take to reach the ground? = (31.6 m/s – 0) 10 m/s2 = 3.16 sec

  11. Let’s look at a coin toss. The coin leaves your hand at 6 m/s and travels upward. The coin speed begins to decrease the moment it leaves your hand. The rate of deceleration is -10 m/s2 The velocity is zero at the top of its trajectory.

  12. Then the coin starts the down trip And begins to accelerate downward Vo = 0 at the rate of + 10 m/s2 When it gets back to your hand its velocity is? The same velocity that it left your hand with 6 m/s Why?

  13. Let’s look at another example of an up down problem. V = 0 m/s - 10 m/s2 down up + 10 m/s2

  14. Example 2 A stone is thrown straight upward with a speed of 20 m/s. a.How long did the trip up take? The entire trip? vo = 20 m/s vf = 0 m/s g = -10 m/s2 = 0 – 20 m/s - 10 m/s2 = 2 s b.How high will the stone go? Using the down trip… d d 0 = (20 m/s)2 + 2(-10 m/s2) d d= 0 + ½ (10m/s2)(2s)2 d = -(20 m/s)2 2(-10 m/s2) = 20 m d= 20m

  15. First, let’s talk about Mr Morris’s boat… . If the boat has a speed of 10 m/s and the current is 5 m/s, what is the resultant velocity of the boat? How long to travel across the river?

  16. If the boat has a speed of 10 m/s and the current is 5 m/s, what is the resultant velocity of the boat? 10 m/s across Θ 5 m/s downstream VR VR = √10 2 + 52 = 11.18 m/s Θ = Tan-1 (5/10) = 26.57º downstream

  17. How long to travel across the 120 m wide river? The time to cross depends on the speed across the river. t = d v = 120 m 10m/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 m/s)(12sec) = 60 m downstream

  18. The BIG Rule; 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…

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

  20. facts about projectiles; 1. Projectiles maintain a constant horizontal velocity (neglecting air resistance) 2. Projectiles always experience a constant vertical acceleration of “g” or 10 m/s2 (neglecting air resistance)

  21. 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.

  22. FACT 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

  23. A plane flying at 115 m/s drops a package from 600m. How far from the drop point will it land? Fact 5. Objects dropped from a moving vehicle have the same horizontal velocity as the moving vehicle.

  24. facts about projectiles; 1. Projectiles maintain a constant horizontal velocity (neglecting air resistance) 2. Projectiles always experience a constant vertical acceleration of “g” or 10 m/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 velocity as the moving vehicle.

  25. Visuals… • Hitting the ground 1st: • Demo: https://www.youtube.com/watch?v=zMF4CD7i3hg • Mythbusters: bullet https://www.youtube.com/watch?v=tF_zv3TCT1U&t=6s Independent components: https://www.youtube.com/watch?v=BLuI118nhzc

  26. Let’s apply these rules to a horizontally thrown projectile… Example 1: 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 = 10 m/s2 For horizontal; dx = ? Vx = 20 m/s a = 0 We can use the vertical to find time… 100m Since horizontal distance … dy = Vot + ½at2 dx = Vxt dx = (20m/s) ? (4.47s) 100m = 0 + ½(10m/s2)t2 dx = 89.4 m t = 4.47 s

  27. Example 2: 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 dx= 47 m vyo = 0 = 3.13 s dy = vy0t + ½ gt2 = (0)(3.13) + ½ (10)(3.13)2 = 48.98 m

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

  29. The angle that a projectile is launched will determine the range and maximum height it obtains. Vy = VR sin θ Vx = VR cos θ

  30. Maximum range is an angle of 47º

  31. Example 1: A golf ball is hit with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Let’s first find the components of the velocity vox = 100 cos 30 = 86.60 m/s vy = 100 sin 30 = 50 m/s VR Vy 30º Vx

  32. a.The time required to reach its maximum height At top vfy = 0vfy = voy + gt 0 = 50 m/s + -10m/s2t t = 5.0 s b.The horizontal range dx Total time T = 2t = 2(5.0) = 10.0 s x = vox t = 86.6(10.0) = 866 m

  33. Example 2: A grasshopper jumps between flowers 1.20 m apart. The grasshopper jumps to a height of 0.40 m, at what velocity and angle does he jump? Horizontal Vertical Need time… which trip to use ? Vyo = dy = .40 m g = 10 m/s2 Vx = dx = 1.2 m t = We will use the down trip. 0.57 s dy = Vot + ½ at2 t = 0.28 s 0.40 m = 0 + ½ (10 m/s2)t2 tdown = 0.28 sec VR 2t = T = 0.57 sec Vy θ Vx

  34. Vfy = Voy + at Horizontal Vertical Vyo = dy = .40 m g = 10 m/s2 Vx = dx = 1.2 m T = Vfy = Voy + 10 m/s2( 0.28s) 0.57 s Vfy = 2.83 m/s t = 0.28 s VR2 = (2.83 m/s)2 + (2.12 m/s)2 VR = 3.54 m/s Vx = dx T = 1.20 m 0.57s Tan θ = 2.83 m/s 2.12 m/s Vx = 2.12 m/s Θ = 53.13º above The ground VR Vy θ Vx

  35. Additional Type II Examples

  36. Horizontal: Vx = 115 m/s dx = ? Vertical: Voy = 0 dy = 600 m a = 10 m/s2 This is a Type II problem we’ve been working… dy = ½ at2 dx = (115 m/s)(10.95s) 600 m= ½ (10m/s2)t2 dx = 1259.25 m t = 10.95 s

  37. 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 = 10 m/s2 22 m = 0 + ½(10m/s2)t2 t = 2.097… s t = ? t = ? = 35.0 m 2.097…s Vx = dx t Vx = 16.69 m/s

  38. What is the vertical velocity just at impact? (Vyf) Horizontal Vertical Voy = 0 dy = 22.0m a = 10 m/s2 Vfy = ? Vx = 16.69m/s dx = 35.0m t = 2.10 s Vfy = Vo + at Vfy = 0 + (10m/s2)(2.09..s) Vfy = 20.98 m/s What is the resultant velocity of the soccer ball at impact? VR2 = (16.69m/s)2+ (20.98m/s)2 = 26.80 m/s VR Vfy Tan θ = 20.98 m/s 16.67 m/s = 51.50º to the ground θ Vx

  39. 3. While standing on a 30 m bridge, a fisherman tosses some unused bait off the bridge. He released it horizontally at 5 m/s. • List the horizontal and vertical givens in a T-chart. B. How long does the bait take to hit the water under the bridge?

  40. 3. While standing on a 30 m bridge, a fisherman tosses some unused bait off the bridge. He released it horizontally at 5 m/s. • C. How fast is the bait traveling vertically when it strikes the water? (Vyf) • D. How fast is it traveling horizontally when it strikes the water? (Vx) • E. What is its range? • F. What is the resultant velocity of the bait as it strikes the water? (This means you use trig and Vx and Vyf as the components to find VR)

  41. 4. A helicopter drops a relief package to some island survivors. The package travels 250 m forward while taking 4 seconds to land after it is released. • List the horizontal and vertical givens in a T-chart.

  42. How high was the helicopter flying when it released the package? B. How fast was the helicopter flying when it released the package? C. How fast was the package falling vertically just before it landed on the island?

  43. D. What was the package’s RESULTANT impact velocity when it landed on the island? E. At what angle was the package’s trajectory with respect to the ground at the instant that it landed? F. If the helicopter pilot does not change his velocity after releasing the package, what will be the position of the helicopter when the package lands?

  44. 5. An Olympic diver springs forward off a 10 m platform. He enters the water 8 m beyond the edge of the platform, • A. List the horizontal and vertical givens in a T-chart. • B. How much time does he spend in the air?

  45. An Olympic diver springs forward off a 10 m platform. He enters the water 8 m beyond the edge of the platform, C. How fast did he spring forward off the edge of the diving tower? (Vx) D. How fast was he traveling vertically when he hit the water? E. What was his resultant velocity when he impacted the water? F. At what angle did he enter the water?

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