# Two-Dimensional Motion and Vectors

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Two-Dimensional Motion and Vectors
Chapter 3

Introduction to Vectors
A scalar is a quantity that can be completely specified by its magnitude A vector is a physical quantity that has both magnitude and direction. Some examples of vector quantities are velocity, acceleration, displacement and forces. Vectors can be added graphically, but when adding vectors you must be sure that they have the same units and that the vectors describe similar quantities. The answer found by adding vectors is called the resultant.

Magnitude of vector → Pythagorean Theorem
Vector operations The magnitude and direction of the resultant of two perpendicular vectors can be calculated using the Pythagorean Theorem and the tangent function. Magnitude of vector → Pythagorean Theorem Direction of vector → Tangent function

Basic Vector Operations Both a magnitude and a direction must be specified for a vector quantity, in contrast to a scalar quantity which can be quantified with just a number. Any number of vector quantities of the same type (i.e., same units) can be combined by basic vector operations.

Set # 9 P/3 A football player runs 15 m down the field, and then turns to the left at an angle of 15º from his original direction and running an additional 10 m before getting tackled Find the magnitude and direction By 10m 15º Bx 15 m

P/5 Set # 9 A person walk’s the path shown below, what’s is the person resultant displacement measured from the starting person 100 m Resultant Vector ? Ax = ______ Ay = ______ Bx = ______ By = ______ Cx = ______ Cy = ______ Dx = ______ Dy = ______ ∑x = ______ ∑y = ______ 300 m 30° 200 m 150 m 60°

1,930 km, 43° south of west 1,850 km, 43° south of west
Example 4: A plane flies from city A to city B. City B is 1,540 km west and 1,160 km south of city A. What is the total displacement and direction of the plane? 1,930 km, 43° south of west 1,850 km, 43° south of west 1,850 km, 37° south of west 1,930 km, 37° south of west

A plane flies from city A to city B
A plane flies from city A to city B. City B is 1,540 km west and 1,160 km south of city A. What is the total displacement and direction of the plane? 1,540 km A 1,160 km B

Example # 5 During a rodeo, a clown runs 8 m north, turns 35º east of north and runs 3.5 m. Then after awaiting for the bull to come near, the clown turns due east and runs 5m to exit the arena. Find the clown’s displacement Resultant Vector 5 m Ax = _____ Ay = _____ Bx = _____ By = _____ Cx = _____ Cy = _____ ∑x = _____ ∑y = _____ 3.5 m 35º 8 m

3.5 m at 19° north of east 6.3 m at 67° north of east
Example 6: A duck waddles 2.5 m east and 6 m north. What is the magnitude and direction of the duck’s displacement with respect to its original position? 3.5 m at 19° north of east 6.3 m at 67° north of east 6.5 m at 67° north of east 6.5 m at 72° north of east

P/1 P/2 A person walks the path shown below:
Physics Set # Oct/2/2007 Student Name: ____________________________________ Class Period: _______ P/1 A person walks the path shown below: Find the person’s resultant displacement measured from the starting point Determine the direction Answers: a)__________ b) ________ P/2 Answers: a)_________ b) _________ 300 m 20º 120 m 150 m 60º 30º 150 m 500 m 250 m 25º 125 m 160 m 15º

Physics Set # 11 Oct/3/2007 Student Name: ____________________________________ Class Period: _______
Mr. C designed play 2244 in football that will send a receiver down the field as explained in the following diagram. Find the displacement of this receiver Find the direction Answers: a) _________ b) _________ P/2 Mr. C designed play 4877 in football that will send a receiver down the field as explained in the following diagram. b) ________ 5 yds 30º 6 yds 7 yds 50º 14 yds 12 yds 4 yds 6 yds 20º 10 yds

Physics Set # 12 Oct/4/2007 Student Name: ____________________________________ Class Period: _______
Mr. C designed play 485 in football that will send a receiver down the field as explained in the following diagram. Find the displacement of this receiver Find the direction Answers: a) _________ b) _________ P/2 Mr. C designed play 4877 in football that will send a receiver down the field as explained in the following diagram. 8 yds 20º 3 yds 40º 4 yds 6 yds 60º 16 yds 11 yds 5 yds 75º 7 yds 40º 15 yds

Physics Set # 13 Oct/4/2007 Student Name: ____________________________________ Class Period: _______
Mr. C designed play 775 in football that will send a receiver down the field as explained in the following diagram. Find the displacement of this receiver Find the direction Answers: a) _________ b) _________ P/2 Mr. C designed play 4877 in football that will send a receiver down the field as explained in the following diagram. 18 yds 40º 4 yds 6 yds 60º 12 yds 8 yds 5 yds 65º 7 yds 40º 12 yds

P/3 P/4 A person walks the path shown below:
Find the person’s resultant displacement measured from the starting point Determine the direction Answers: a)__________ b) ________ P/4 Answers: a)_________ b) _________ 250 m 20° 20º 50 m 60 m 60º 30º 500 m 180 m 250 m 25º 125 m 160 m 15º

Vectors The study of vectors allows for breaking a single vector into components: the vertical component the horizontal component. By breaking a single vector into two components, or resolving it into its components, an object’s motion can sometimes be described more conveniently in terms of directions, such as north to south, or up and down. vi vy vx

Essay: Wernher von Braun 2 pages/ 12 font Chunk Paragraph: topic sentence + 2 concrete details +4 commentaries concluding sentence The early childhood Family His achievements in Germany His involvement in the Nazi Party How did he ended up in U.S.A.? His achievements in U.S.A.

Projectile Motion

Projectile Motion A projectile is any object upon which the only force acting is gravity. Projectiles travel with a parabolic trajectory due to the influence of gravity, There are no horizontal forces acting upon projectiles and thus no horizontal acceleration, The horizontal velocity of a projectile is constant (never changing in value), There is a vertical acceleration caused by gravity; its value is -9.81 m/s², down. The vertical velocity of a projectile changes by – 9.81 m/s each second. The horizontal motion of a projectile is independent of its vertical motion.

Projectile Motion A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object which is thrown vertically upwards is also a projectile (provided that the influence of air resistance is negligible). An object which is thrown upwards at an angle is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object which once projected continues in motion by its own inertia and is influenced only by the downward force of gravity.

Projectile Motion By definition, a projectile has only one force acting upon - the force of gravity. If there were any other force acting upon an object, then that object would not be a projectile. Thus, the free-body diagram of a projectile would show a single force acting downwards and labeled "force of gravity" (or simply Fg). This is to say that regardless of whether a projectile is moving downwards, upwards, upwards and rightwards, or downwards and leftwards, the free-body diagram of the projectile is still as depicted in the following diagram

Projectile Motion

Projectile Motion

Projectile Motion

Projectile Motion Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward and/or downward they are also moving horizontally. There are the two components of the projectile's motion - horizontal and vertical motion. And since "perpendicular components of motion are independent of each other," these two components of motion can (and must) be discussed separately. The goal of this part of the lesson is to discuss the horizontal and vertical components of a projectile's motion; specific attention will be given to the presence/absence of forces, accelerations, and velocity.

Projectile Motion

Projectile Motion In this course, we are going to study 2 different situations about projectiles: 1. Projectiles launched horizontally 2. Projectiles launched at an angle θ

Projectiles launched horizontally
Horizontal motion of a projectile ∆x = vx ∙∆t vx = vi,x = constant Vertical Motion of a projectile ∆y = ½ a(∆t)² vy,f = a ∆t vy,f ² = 2 a∆y

Projectiles launched at an angle
v vy vx,i = vi cos Ө vy,i = vi sin Ө vx Ө Horizontal motion of a projectile ∆x = vi ∙cos θ∙∆t vx =vi ∙cosθ = constant Vertical Motion of a projectile ∆y = vi∙sinθ ∙∆t +½ a(∆t)² vy,f = vi∙sin θ + a∙∆t vy,f² = vi² (sin θ)² +2 a∙∆y

HORIZONTAL MOTION OF A PROJECTILE VERTICAL MOTION OF A PROJECTILE
Example #1 The New York Yankees, in their quest for an unprecedented 27th. World Series Championship, have an extraordinary pitching rotation. One of our best pitchers is Mike Mussina,aka Moose, who is capable of throwing his fastball at 92 mph. If he pitched to the home-plate (horizontally) (60 ft away), how far will the ball fall vertically by the time it reaches home plate?. HORIZONTAL MOTION OF A PROJECTILE ∆x = vix∙∆t vix = constant vix = vx VERTICAL MOTION OF A PROJECTILE ∆y = a (∆t)² 2

∆y = ½ a∙(∆t)² = ½ (-9.81 m/s²) (0.44 s)² = - 0.95 m
Example # 1 Moose, who is capable of throwing his fastball at 92 mph. If he pitched to the home-plate (horizontally) (60 ft away), how far will the ball fall vertically by the time it reaches home plate?. Given: vix = 92 mi/h → vix = 41 m/s ∆x = 60 ft → ∆x = m Unknowns: ∆t =?, ∆y=? Solution: First find the time that it takes for the baseball to reach home plate: ∆x = vix∙∆t ∆t = ∆x/vix ∆t = m / 41 m/s = 0.44 s Now find the vertical fall: ∆y = ½ a∙(∆t)² = ½ (-9.81 m/s²) (0.44 s)² = m

Given: ∆y = -120 m a= -9.81 m/s2 vx = 40 m/s Unknowns: ∆t, ∆x
Example # 2 A plane wants to drop a cargo package into a drop area. The plane is 120 m above the ground, and traveling at a velocity of 40 m/s in the positive x direction. Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area. Given: ∆y = -120 m a= m/s vx = 40 m/s Unknowns: ∆t, ∆x 120 m ∆x Drop Zone

Given: ∆y = -120 m a= -9.81 m/s2 vi,x = 40 m/s vi,y = 0 m/s
Example 2: A plane wants to drop a cargo package into a drop area. The plane is 120 m above the ground, and traveling at a velocity of 40 m/s in the positive x direction. Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area. Given: ∆y = -120 m a= m/s vi,x = 40 m/s vi,y = 0 m/s Unknowns: ∆t, ∆x Solution: ∆y = ½ a ∆t² → ∆t² = 2∆y/a ∆t² = (2) (-120 m)/ ( m/s²) ∆t² = → ∆t = √ 24.64 ∆t = 4.96 s ∆x = vi,x ∙∆t ∆x = (40 m/s) (4.96 s) ∆x = m

Example # 3 People in movies often jump from buildings into pools
Example # 3 People in movies often jump from buildings into pools. If a person jumps from a 10th. floor(30 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump? 30 m 5 m Given: Δy = - 30 m Δx = 5 m a = m/s²

Example 4: A zookeeper finds an escaped monkey hanging from a light pole. Aiming his tranquilizer gun at the monkey, the zookeeper kneels 12 m from the light pole which is 6 m high. The tip of his gun is 1 m above the ground. The monkey tries to trick the zookeeper by dropping a banana, then continues to hold onto the light pole. At the moment the monkey releases the banana, the zookeeper shoots. If the tranquilizer dart travels at 50 m/s, will the dart hit the monkey or the banana? 6 m 1m 12 m

Steps 1. Find the angle : Θ = tan‾¹ (5/12) = 23º
2. Find the time it takes to reach the target: ∆x = vi ∙cos θ∙∆t ∆t = ∆x / vi ∙cos θ ∆t =

Given: vi = 1,700 m/s Θ = 30° a = -9.81 m/s² Unknowns: ∆t, ∆x, ∆y
Example 3: A ball is fired from the ground with an initial speed of 1,700 m/s at an initial angle of 30° to the horizontal. Neglecting air resistance, find: a) The ball’s horizontal range b) The amount of time the ball was in motion Given: vi = 1,700 m/s Θ = 30° a = m/s² Unknowns: ∆t, ∆x, ∆y Steps: ∆x = vi cos Θ ∆t → ∆x = (1,700 m/s) cos 30°∙ ∆t ∆x = 1, m/s ∆t ∆y = vi sin Θ ∆t + ½ a (∆t)² → ∆y = (1,700 m/s) sin 30° ∆t + ½ ( m/s²) ∆t² ∆y = 850 m/s ∆t – 4.91 ∆t² ∆y = 0 (because it ends at the same vertical position) 4.91 ∆t²- 850 ∆t = 0 4.91 ∆t- 850 = 0 ∆t = 173 s ∆x = (1, m/s) (173 s) ∆x = 254, m

Golf and Physics When you play golf, you have a variety of golf clubs to hit for different distances. Most people use the driver to tee off on the par 4’s, and then a 7 iron from the middle of the fairway. The driver has typically an angle on 10˚, and the 7 iron an angle of 35˚, Calculate the horizontal and vertical displacement of a golfer that hits golf balls at an average initial velocity of 120 mi/h for the initial 10 s after impact. www,explore science.com Y axis Height X axis (0,0) Time (t,0)

Driver 10˚ vi = 120 mi/h (53.63 m/s) ∆t ∆x = vi cosθ ∆t
∆y = vi sin θ ∆t + ½ a ∆t ² 1s 52.82 m 4.41 m 2s m m 3s m 4s m 5s m 6s m 7s m 8s m 9s m 10s m

7 iron 35˚ vi = 120 mi/h (53.63 m/s) ∆t ∆x = vi cosθ ∆t
∆y = vi sin θ ∆t + ½ a ∆t ² 1s 43.93 m 25.86 m 2s 87.86 m 41.90 m 3s m 48.14 m 4s m 44.56 m 5s m 31.18 m 6s m (241 yd) 7.99 m 7s m m 8s m 9s m 10s m

Something’s wrong??? The average displacement for a 7 iron is 150 yds!! Calculate the angle for: Vi = 120 mi/h → m/s ∆x = 150 yds → m ∆t = 6.3s ? θ = ?????

BONUS POINTS The great pyramid of Giza is approximately 150 m high and has a square base approximately 230 m on a side. What is the approximate area of a horizontal cross section of the pyramid taken 50 m above its base? (Determine this by using 3 different methods!!) 5,880 m² 11,760 m² 23,510 m² 35,270 m²

A F B E C G D

CD = 230 m AG = 150 m GF = 50 m A F B E C G D

CD = 230 m AG = 150 m GF = 50 m AF = 100 m A AG = AF CD BE BE = m BE² = 23,511 m² 100 m F B E 50 m C G D 230 m

BONUS POINTS Using a protractor, a student measures the sum of the interior angles in a triangle and obtains 176°. What is the percent error of this measurement? 0.04% 2.22% 2.27% 4.00%

BONUS POINTS For what value of c will the function below have exactly one vertical asymptote? y = ____4______ x² – cx + 9 4 6 9

Comparing a Driver and a 7 iron
∆y (m) 100 m 75 m 100 50 25 ∆x (m)

Projectile Motion Example # 3 From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. The bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.60 m, and x = 6.8 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window

Example 4: A golfer practices driving balls of a cliff and into water below. The cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15°. How far does the ball travel horizontally before hitting the water? 15 m

For ax² + bx + c = 0 the value of x is given by:

Summer School Physics. Set # 9
Summer School Physics Set # June/16/2008 Student Name: _____________________________________ Class Period: ______ P/1 A golfer practices driving balls of a cliff and into water below. The cliff is 12 m above the water. If the golf ball is launched at 61 m/s at an angle of 25°. How far does the ball travel horizontally before hitting the water? Answer: ___________ P/2 A place kicker must kick a football from a point 36 m (about 40 yd) from the goal, and the ball must clear the crossbar which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20 m/s at an angle of 53° to the horizontal. A. By how much does the ball clear of fall short of clearing the crossbar? B. Does the ball approach the crossbar while still rising or while falling Answers: A = ___________ B = ___________ P/3 A daredevil jumps a canon 12 m wide. To do so, he drives a car up a 15° incline. A) What minimum speed must he achieve to clear the canyon? B) If the daredevil jumps at this minimum speed, what will be his speed when he reaches the other side?

Example 4: A golfer practices driving balls of a cliff and into water below. The cliff is 15 m from the water. If the golf ball is launched at 51 m/s at an angle of 15°. How far does the ball travel horizontally before hitting the water? Given: ∆y = - 15 m, v = 51 m/s, Ө = 15° Solution: 1. Find the time that it takes the golf ball to hit the water 2. Find the horizontal displacement To find the time: ∆y = vi sinӨ ∆t + ½ a ∆t² - 15 =(51 m/s) sin15°∆t - ½ (9.81) ∆t² - 15 = 13.2 ∆t – 4.91 ∆t² 4.91 ∆t²  13.2 ∆t – 15 = 0 ∆t = - b ±√b²  4 ac 2 a ∆t = ± 21.65 ∆t = 3.55 s ∆x = vi cosӨ∙∆t ∆x = m

Example # 5 A place kicker must kick a football from a point 36 m (about 40 yd) from the goal, and the ball must clear the crossbar which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20 m/s at an angle of 53° to the horizontal. A. By how much does the ball clear of fall short of clearing the crossbar? B. Does the ball approach the crossbar while still rising or while falling? ∆y = 3.05 m Ө = 53° ∆x = 36 m

Given: ∆x = 36 m,, Ө = 53°, a = - 9.81 m/s², vi = 20 m/s ∆y = 3.05 m ?
A place kicker must kick a football from a point 36 m (about 40 yd) from the goal, and the ball must clear the crossbar which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20 m/s at an angle of 53° to the horizontal. Given: ∆x = 36 m,, Ө = 53°, a = m/s², vi = 20 m/s ∆y = 3.05 m ? Unknowns: ∆t Solution: First find the time that it takes for the ball to reach the goal post, and then determine how high will the ball be at that time to see if it clears the goal post. ∆x = vi cosӨ ∆t → 36 m = (20 m/s) (cos 53°) ∆t 36 m = (12.04 m/s) ∙∆t ∆t = 36 m/ m/s 2.99 s

Example # 6: A daredevil jumps a canon 12 m wide
Example # 6: A daredevil jumps a canon 12 m wide. To do so, he drives a car up a 15° incline. A) What minimum speed must he achieve to clear the canyon? B) If the daredevil jumps at this minimum speed, what will be his speed when he reaches the other side?

Physics Set # October 30, 2007 Student Name: ____________________________________ Class Period: _______ P/1 If you throw a baseball horizontally at 85 mph to home plate (60 ft away), how much will the baseball drop, once it reaches home plate (neglecting air resistance)? Answer: ____________ P/2 If you throw a baseball horizontally at 75 mph to home plate (60 ft away), how much will the baseball drop, once it reaches home plate (neglecting air resistance)? P/3 A plane wants to drop a cargo package into a drop area. The plane is 150 m above the ground, and traveling at a velocity of 50 m/s in the positive x direction. Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area. Answer: ___________ P/4 A plane wants to drop a cargo package into a drop area. The plane is 250 m above the ground, and traveling at a velocity of 80 m/s in the positive x direction. Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area.

Physics Set # November 2, 2007 Student Name: ____________________________________ Class Period: _______ P/1 A ball is fired from the ground with an initial speed of 500 m/s at an initial angle of 40° to the horizontal. Neglecting air resistance, find: a) The ball’s horizontal range Answer: ___________ b) The amount of time the ball was in motion Answer: ___________ c) The maximum height that the ball reaches Answer: ___________ P/2 A ball is fired from the ground with an initial speed of 400 m/s at an initial angle of 60° to the horizontal. Neglecting air resistance, find: P/3 A golfer practices driving balls of a cliff and into water below. The cliff is 15 m from the water. If the golf ball is launched at 60 m/s at an angle of 10°. a) How far does the ball travel horizontally before hitting the water? Answer: __________ b) What was the maximum height of the ball while in the air?

Steps: ∆y = viy ∆t + ½ a (∆t)² viy = 0 ∆y = ½ a (∆t)² ________ ∆t = √ 2 ∆y a ∆t = 4.95 s ∆x = Vx∆t = (40 m/s) (4,95 s) = 198 m

Bonus Points A fifth-grade class is using pattern blocks in the shape of congruent equilateral triangles to devise and solve problems involving fractions. One group devises the problem illustrated below. Given that the sum of Shapes A and B represents 5/8, which of the following represent 1 ¼? A)

Projectile Motion Physics

Driver 10˚ vi = 120 mi/h ∆t ∆x = vi cosθ ∆t
∆y = vi sin θ ∆t + ½ a ∆t ² 1s 2s 3s 4s 5s 6s 7s 8s 9s 10s

7 iron 35˚ vi = 120 mi/h ∆t ∆x = vi cosθ ∆t
∆y = vi sin θ ∆t + ½ a ∆t ² 1s 2s 3s 4s 5s 6s 7s 8s 9s 10s