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How to do kinematics in 2-D Remember the ball shot at the monkey. Motion, force and accelerations in the X direction do not affect Y motion. And vice versa: motion, force and accelerations in the Y direction do not affect X motion. So motion needs to be treated as two separate problems: 1 problem along the X axis and 1 problem in the Y axis and the two problems DO NOT AFFECT each other.

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How to do kinematics in 2-D 1) Step 1: List x and y variables separately. Use geometry or trig to separate into components (use cos, sine, tangent). (or pick a new frame of reference to make the cosine and sine go away. Example, the escalator problem. Slant the angle of the X axis.) 2) Treat x and y like two different problems. 3) Solve x and y. 4) Recombine x and y vectors with geometry or trig to find the Resultant Vector. (Find the resultants or total vectors)

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Vector Review, Displacement Δx = “horizontal displacement” Δy = “vertical displacement” => “total displacement” = Magnitude and Angle

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v x = “horizontal component of velocity” v x = v total cosΘ v y = “vertical component of velocity” v y = v total sinΘ v total = Velocity

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Angle Review Trig: For any triangle: From SOHCAHTOA Θ disp = cos -1 (X/H) Θ vel = cos -1 (V X /V total ) Or Θ disp = tan -1 (Y/X) Θ vel = tan -1 (V Y /V X )

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Acceleration a x = “horizontal component of acceleration” a x = a total cosΘ a y = “vertical component of acceleration” a y = a total sinΘ a total = Find the angle also.

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2-D Kinematic Equations Δx = x i + v ix t+ ½ a x t 2 v fx 2 = v ix 2 + 2a x Δx Δv x = a x t Δy = y i + v iy t+ ½ a y t 2 v fy 2 = v iy 2 + 2a y Δy Δv y = a y t Note: t is the only variable common to both dimensions.

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2-D Kinematic Equations Δx = x i + v ix t+ ½ a x t 2 v fx 2 = v ix 2 + 2a x Δx Δv x = a x t Δy = y i + v iy t+ ½ a y t 2 v fy 2 = v iy 2 + 2a y Δy Δv y = a y t Freefall: This is a special case where a x = 0 a y = -9.80 m/s 2 = Gravity Assume: No air resistance, no friction, no forces other than gravity.

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3.2 Equations of Kinematics in Two Dimensions The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa.

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3.2 Equations of Kinematics in Two Dimensions Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2. a)Total initial velocity and acceleration Find, after 7.0 s. b) Δx and v fx, c) Δy and v fy, and d)total final velocity

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3.2 Equations of Kinematics in Two Dimensions

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2 2 2

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Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2. Find (a) x and v x, (b) y and v y, and (c) the final velocity of the spacecraft at time 7.0 s. xaxax vxvx v ox t ?+24.0 m/s 2 ?+22 m/s7.0 s yayay vyvy v oy t ?+12.0 m/s 2 ?+14 m/s7.0 s

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3.2 Equations of Kinematics in Two Dimensions xaxax vxvx v ox t ?+24.0 m/s 2 ?+22 m/s7.0 s

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3.2 Equations of Kinematics in Two Dimensions yayay vyvy v oy t ?+12.0 m/s 2 ?+14 m/s7.0 s

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3.2 Equations of Kinematics in Two Dimensions

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Projectile motion Most common 2D kinematics type. An object gets “thrown” at some angle.

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Projectile Motion (Type of Freefall) An object may move in both the x and y directions simultaneously –It moves in two dimensions The form of two dimensional motion we will deal with is an important special case called projectile motion

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Assumptions of Projectile Motion We make a few assumptions: –Object has no propulsion of its own. –The only acceleration on the object is gravity. –We ignore air friction –We ignore the rotation of the earth (Question: How do we ignore the earth’s rotation???)

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Assumptions of Projectile Motion –We may ignore the rotation of the earth (How do we ignore the earth???) ANS: Put our frame of reference on the surface of the earth so that the frame of reference rotates as the earth rotates. i.e., we use a moving origin, fixed to the surface of the earth.

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Free fall problems, pg 1/2 y-direction – v iy = v itotal sinΘ –Free fall problem a y = -g = -9.80 m/s 2 –Take the positive direction as upward –Uniformly accelerated motion, so the motion equations all hold

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Free fall problems, pg 2/2 x-direction – v ix = v itotal cosΘ –No acceleration a x = 0 v ix = v fx Δx = v ix t –Usually: Right X is positive, left X is negative, or East is positive, west is negative.

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Free fall problems What does the path of a projectile (i.e. a launch object in freefall) look like? (Launch Monkey ball demo here at 3 different angles. Have class guess.) Answer on next slide.

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The result A parabola

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See, it’s a parabola

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Time of Travel Time of flight does not depend on v x or Δx at all. Time of flight is only determined by the Y motion.

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No matter what angle you start at, it’s a parabola. The maximum range occurs at a projection angle of 45 o Complementary values of the initial angle result in the same range –The heights will be different

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Example A Falling Care Package The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the care package to hit the ground.

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3.3 Projectile Motion yayay vyvy v oy t -1050 m-9.80 m/s 2 0 m/s?

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3.3 Projectile Motion yayay vyvy v oy t -1050 m-9.80 m/s 2 0 m/s?

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What is the final velocity vector before it hits the ground?

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Does V x change? Does V y change? Does V total change?

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yayay vyvy v oy t -1050 m-9.80 m/s 2 ?0 m/s14.6 s

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yayay vyvy v oy t -1050 m-9.80 m/s 2 ?0 m/s14.6 s

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3.3 Projectile Motion

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Conceptual I Shot a Bullet into the Air... Suppose you are driving a convertible with the top down. The car is moving to the right at constant velocity. You point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle?

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Conceptual I Shot a Bullet into the Air... Suppose you are driving a convertible with the top down. The car is moving to the right at constant velocity. You point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle?

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3.3 Projectile Motion Typical Projectile Solution Pattern: 1.Calculate something in y direction. Use it to: 2. Find the time of flight. 3. Find the X displacement.

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3.3 Projectile Motion The Height of a Kickoff A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.

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3.3 Projectile Motion The Height of a Kickoff A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, find total time of flight.

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3.3 Projectile Motion The Height of a Kickoff A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, Find the range.

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3.3 Projectile Motion What did we assume that we have not yet stated???

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3.3 Projectile Motion What did we assume that we have not yet stated??? Ball starts at 0 feet high. Close enough to count! That’s an engineering answer.

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3.3 Projectile Motion

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yayay vyvy v oy t ?-9.80 m/s 2 014 m/s

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3.3 Projectile Motion. Find height or y max first. yayay vyvy v oy t ?-9.80 m/s 2 014 m/s

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3.3 Projectile Motion Example 7 The Time of Flight of a Kickoff What is the time of flight between kickoff and landing?

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3.3 Projectile Motion yayay vyvy v oy t 0-9.80 m/s 2 14 m/s?

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3.3 Projectile Motion yayay vyvy v oy t 0-9.80 m/s 2 14 m/s? / t and Clear Fractions

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3.3 Projectile Motion Example 8 The Range of a Kickoff Calculate the range R of the projectile.

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Non-Symmetrical Projectile Motion Follow the general rules for projectile motion Break the y-direction into parts –up and down –symmetrical only back to initial height.

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Non-Symmetrical Projectile Motion Which part of this math is symmetrical and which is not?

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Note to teacher Print out rest of these pages and give to the class. The powerpoint handout of these is already created as a separate file to be printed out with the answer key placed on the teacher’s desk.

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Conceptual Two Ways to Throw a Stone From the top of a cliff, a person throws two stones. The stones have identical initial speeds, but stone 1 is thrown downward at some angle below the horizontal and stone 2 is thrown up at the same angle above the horizontal. Neglecting air resistance, which stone, if either, strikes the water with greater velocity?

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Non-Symmetrical Projectile Motion Find the time of flight and the range of this bowling ball we’re throwing off the roof.

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2. An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. a. At what horizontal distance from the skiers should the supplies be dropped? b. Find the magnitude of the velocity of the box as it reaches the ground.

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2. An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. a. At what horizontal distance from the skiers should the supplies be dropped? b. Find the magnitude of the velocity of the box as it reaches the ground. v x = 80 m/s y = 250 m a= -9.80 m/s 2 = 7.14 s x = v x t = (80)(7.14) = 571 m v x = 80 m/s v y = gt = (9.8)(7.14) = 70 m/s = 106 m/s

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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?

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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

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4. 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

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4. 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 = - 9.8 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) = 692.8 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) = - 28.4 m/s V = (86.6 2 + -28.4 2 ) 1/2 V = 91.1 m/s θ = Inv Tan (-28.4/86.6) θ = -18.2

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b. The time required to reach its maximum height c. The horizontal distance (range)

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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) = 883.7 m

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5. 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 was it released?

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5. 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

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b. What was the maximum height achieved by the ball?

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y = v oy t +½gt 2 = (15)(sin 40.8º)(1) + ½ (-9.8)(1) 2 = 4.9 m

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