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.

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)

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

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

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 )

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.

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.

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 = m/s 2 = Gravity Assume: No air resistance, no friction, no forces other than gravity.

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.

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

3.2 Equations of Kinematics in Two Dimensions

2 2 2

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

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

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

3.2 Equations of Kinematics in Two Dimensions

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

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

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???)

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.

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

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.

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.

The result A parabola

See, it’s a parabola

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.

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

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.

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

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

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

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

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

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

3.3 Projectile Motion

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?

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?

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.

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.

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.

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.

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

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.

3.3 Projectile Motion

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

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

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

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

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

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

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.

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

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.

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?

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

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.

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= 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

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?

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

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

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 = 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 V = ( ) 1/2 V = 91.1 m/s θ = Inv Tan (-28.4/86.6) θ = 

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

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

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?

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

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

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