Projectile HW #1: Handout

Slides:



Advertisements
Similar presentations
Motion in One Dimension
Advertisements

Introduction to 2-Dimensional Motion
Introduction to 2-Dimensional Motion
Section 3-5: Projectile Motion
Two-Dimensional Motion and Vectors
Chap 3 :Kinematics in 2D, 3D and Projectile Motion HW4: Chap.3:Pb.14,Pb.57, Pb.87 Chap 4:Pb.3, Pb.4, Pb.12, Pb.27, Pb. 37 Due Friday 26.
Chapter 6B – Projectile Motion
PLAY Physics Con-Seal From RegentsEarth.com.
Projectile Motion Chapter 3.
Aim: How can we approach projectile problems?
CHAPTER 3 PROJECTILE MOTION. North South EastWest positive x positive y negative x negative y VECTORS.
2-D Motion Because life is not in 1-D. General Solving 2-D Problems  Resolve all vectors into components  x-component  Y-component  Work the problem.
Projectile Motion Physics 6A Prepared by Vince Zaccone
Chapter 5 Projectile motion
Introduction to 2-Dimensional Motion. 2-Dimensional Motion Definition: motion that occurs with both x and y components. Each dimension of the motion can.
Chapter 2 Preview Objectives One Dimensional Motion Displacement
Kinematics in Two Dimensions Chapter 3. Expectations After Chapter 3, students will:  generalize the concepts of displacement, velocity, and acceleration.
You are going 25 m/s North on I-35. You see a cop parked on the side of the road. What is his velocity related to you. A.25 m/s South B.25 m/s North C.0.
AIM: How can we describe the path of an object fired horizontally from a height above the ground? DO NOW: A ball rolls off a table top with an initial.
Do now A B + = ? The wrong diagrams Draw the right diagram for A + B.
Lecture 4: More kinematics. Displacement and change in displacement Position vector points from the origin to a location. The displacement vector points.
Kinematics in 2-Dimensional Motions. 2-Dimensional Motion Definition: motion that occurs with both x and y components. Example: Playing pool. Throwing.
Kinematics in Two Dimensions Chapter Displacement, Velocity, and Acceleration.
Projectiles.
3-7 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.
Return to Table of Contents Acceleration What is constant speed? If the speed of an object does not change, the object is traveling at a constant speed.
PHYSICS: Vectors and Projectile Motion. Today’s Goals Students will: 1.Be able to describe the difference between a vector and a scalar. 2.Be able to.
Physics. Good News/Bad News: These are the same formulas we used for linear motion. Do you know them? If the answer is “NO”, then get familiar with them.
Physics pre-AP. Equations of motion : We assume NO AIR RESISTANCE! (Welcome to “Physicsland”), therefore… The path of a projectile is a parabola. Horizontal.
Physics Honors. Good News/Bad News: These are the same formulas we used for linear motion. Do you know them? If the answer is “NO”, then memorize them.
Kinematics in 2-D Concept Map
College Physics, 7th Edition
Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 3 Motion in 2 Dimensions.
2 – Dimensional Kinematics PROJECTILE MOTION (Right-click to pause presentation at any time)
Copyright Sautter Motion in Two Dimension - Projectiles Projectile motion involves object that move up or down and right or left simultaneously.
Copyright Sautter General Problem Solving Steps (1) Read the problem more than once (three of four times is preferable) (2) Decide what is to be.
Chapter 4 Two-Dimensional Kinematics. Units of Chapter 4 Motion in Two Dimensions Projectile Motion: Basic Equations Zero Launch Angle General Launch.
Projectile Motion Examples. Example 3-6: Driving off a cliff!! y is positive upward, y 0 = 0 at top. Also v y0 = 0 v x = v x0 = ? v y = -gt x = v x0 t,
Kinematics in Two Dimensions. Section 1: Adding Vectors Graphically.
Chapter 3 Kinematics in Two Dimensions. 3.1 – d, v, & a A bullet is fired horizontally. A second bullet is dropped at the same time and at from the same.
MOTION IN ONE DIMENSION AVERAGE / INSTANTANEOUS SPEED POSITION AND DISPLACEMENT AVERAGE / INSTANTANEOUS VELOCITY AVERAGE / INSTANTANEOUS ACCELERATION.
PROJECTILE MOTION. Relevant Physics: The Independence of the Vertical and Horizontal directions means that a projectile motion problem consists of two.
CHAPTER 6 MOTION IN 2 DIMENSIONS.
Motion in Two Dimensions. Projectile Motion: the motion of a particle that is projected or launched and only accelerated by gravity. cp: 5.
Motion in Two Dimensions. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.
CHAPTER 3 MOTION IN A PLANE
Motion in Two Dimensions. (Ignore any effects from air resistance) A pickup is moving with a constant velocity and a hunter is sitting in the back pointing.
Position and Displacement If we want to locate a point in space we can use a position vector that extends from a reference point to the location we are.
2D Motion 2015 cjcb Angled Projectile Motion. RECAP: What is Projectile Motion? Projectile motion is the motion of an object that is being thrown or launched.
Newton laws Application Projectile Motion First of All, What is 2-D Motion? Before, we talked about motion in one dimension (just the x axis) Now we.
Continued Projectile Motion Practice 11/11/2013. Seed Question Six rocks with different masses are thrown straight upward from the same height at the.
Kinematics in Two Dimensions. I. Displacement, Velocity, and Acceleration A. Displacement- distance traveled from starting point to final destination.
Introduction to 2D Motion
Projectile Motion Introduction Horizontal launch.
AP PHYSICS Chapter 3 2-D Kinematics. 2-D MOTION The overriding principle for 2-Dimensional problems is that the motion can be resolved using vectors in.
Physics for Dentistry and Medicine students Physics for Dentistry and Medicine students PHYS 145 Text book Physics; John D. Cutnell and Kenneth W. Johnson;
Kinematics in Two Dimensions Vectors
Projectile Motion Physics Level.
Physics Support Materials Higher Mechanics and Properties of Matter
3-7 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. Figure Caption:
Projectile motion.
Unit 3: Projectile & 2D Motion
9.8: Modeling Motion Using Parametric Equations
Projectile Motion Physics Honors.
Projectile Motion AP Physics C.
9.8: Modeling Motion Using Parametric Equations
Vectors add direction to a magnitude.
Projectile Motion Physics Honors.
College Physics, 7th Edition
Presentation transcript:

Projectile HW #1: Handout Projectile Motion Projectile HW #1: Handout

Projectile Motion: Introduction: Projectile motion refers to freefall motion in ______ dimensions. The motion of the object will have a ______________ component and a _______________ component. There is a constant acceleration due to gravity, , points downwards only. Here the g refers to: two horizontal vertical downwards downwards There are two coordinates to describe projectile motion. The _____ coordinate and component refer to _____________ motion and the _____ coordinate and component refer to _____________ motion. x horizontal y vertical We need to define the same concepts used in Ch. 2 for our study of two dimensional motion.

Position: An object must be given a location in space Position: An object must be given a location in space. A two dimensional coordinate system is used: +y = up The object’s location can be described in relation to the origin. The origin can be chosen to be any place convenient. The position can be represented by a vector whose coordinates are (x,y). +x = right, or any other horizontal direction The coordinates (x,y) also represent the x and y components of the vector position.

Displacement: The displacement of an object is the ____________ of ____________ of an object. The displacement is a ____________! The displacement is represented as: change position vector +y Initial position +x Final position

Average velocity: The average velocity of an object represents the __________ at which position ____________. rate changes The average velocity is the displacement of the object divided by the ______________ time. elapsed To simplify the equations, we will always take the initial time to be __________ seconds. zero With this change, the ______ time always equals the ________ time. final elapsed

Instantaneous velocity: The instantaneous velocity of an object represents the __________ at one ____________ of time. The instantaneous velocity is represented by the letter v, and it is also a vector. velocity instant Vx θ To find the resultant, apply Pythagoras and tan-1 θ Vy

Average acceleration: The average acceleration of an object represents the __________ at which velocity ____________. rate changes velocity The average acceleration is the change of the ___________ of the object divided by the ______________ time. elapsed Instantaneous acceleration: The instantaneous acceleration of an object represents the _____________ at one ____________ of time. The instantaneous acceleration is represented by the letter a, and it is also a vector. acceleration instant

Uniformly Accelerated Motion: This kind of motion has a constant ________________. Since the acceleration is constant, the _____________ and the _________________ acceleration are equal. acceleration average instantaneous Freefall: The acceleration of an object will be due to gravity only. Gravity pulls with a constant acceleration towards the ground, or _____________ downward. The acceleration can be written as a vector as follows: vertically As before, g just represents the numeric value of the acceleration of gravity. The – sign shows the direction points downwards!

2 – dimensional motion difficulty: The motion of a projectile in two dimensions is quite complex! The way to solve these problems is to resolve all the motion into components.

All the motion completely separates into components All the motion completely separates into components. The motion in the x direction is independent of the motion in the y direction. By separating the motion into x and y components, the motions become independent of one another. The motion in the x – direction is not affected by the motion in the y – direction.

Equations of Motion: The motion of a projectile in each dimension can be represented by the uniform motion equations from Ch. 2. Apply the equations to each coordinate axis separately. The general equations for uniform motion from Ch. 2 are:

Apply these equations to each coordinate axis Apply these equations to each coordinate axis. For the x – direction, the acceleration is _________. The equations from Ch. 2 can be adapted to 2 –dimensional motion by adding subscripts to the variables that are vectors. The subscripts give the component name: zero The acceleration in the x – direction, ax, is zero. Dx is the horizontal displacement. vox is the initial velocity in the x direction. vx is the final velocity. The x component of the velocity is constant and distance is rate times time.

ay = – g For the y – direction, the acceleration is _________. Substitute y for x and get the equations of motion for the vertical part of the motion:

The motion in the x – direction and the y – direction are independent of one another. The motion of a projectile is that of a constant velocity in the x – direction and a uniformly accelerated motion in the y – direction.

To solve projectile problems, just follow the basic guidelines given in Ch. 2. Read the problem, draw a picture, write down what is given and unknown, choose equations, and solve. Example #1: A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach, as shown below. a. How long after being released does the stone strike the beach below the cliff? b. How far from the base of the cliff does the stone land? c. With what speed and angle of impact does the stone land?

vox a. Use the y – direction to find the time to fall. Dy = – 50.0 m voy = 0

solve for t: b. Use the x – direction to find the horizontal distance.

c. Solve for each component of the final velocity just before impact. For the x – direction: For the y – direction: Note: The minus sign on vy is to show it points downwards!

vx = +18.0 m/s Use Pythagorean theorem: q vy = – 31.3 m/s v Use inverse tangent: below horizontal

Example #2: A ball is launched horizontally from a height h above the ground. At the same moment, an identical ball is dropped from rest from the same height. Which ball will hit the ground first? Solution: Both balls hit the ground at the same time. For the ball launched horizontally, the motion in the x – direction and the motion in the y – direction separate. The motion in the y – direction only depends on the y – direction, and is independent of the x – direction. That means the ball dropped straight down and the ball launched horizontally have the same vertical motion. They should arrive at the bottom at the same time with the same vertical speed. Video demonstration:

Example #3: A hunter aims his banana cannon directly at a monkey hanging on a branch. If the monkey lets go of the branch at the moment the cannon is fired, will the monkey catch the banana? Both banana and monkey are accelerated equally gravity. No matter how slow the banana is fired, or how far the monkey falls, the two will always make contact. This can be shown a with a couple of different scenarios. First what would happen if gravity were “turned off”?

If gravity were “off”, the banana would travel directly to the monkey.

If gravity is restored, and the cannon points directly to the monkey, the banana will still arrive to the monkey. Both fall the same distance due to gravity as they move. Here is a high projectile speed example:

Here is a low projectile speed example:

Here is a high projectile speed aimed too high:

Example #4: A student stands at the edge of a building and throws a stone horizontally over the edge with a speed of 12.0 m/s. The stone lands 2.41 seconds after it is thrown. a. How tall is the building? b. How far from the base of the building does the ball land? c. With what speed and angle of impact does the stone land? Look for an equation with Dy and t and solve. vox = 12.0 m/s voy = 0 m/s yo = h = ? y = 0 Dy = 0 – h

b. Solve for the x – direction displacement: c. Solve for each component of the final velocity just before impact. For the x – direction: For the y – direction: The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = +12.0 m/s Use Pythagorean theorem: q vy = – 23.6 m/s v Use inverse tangent: below horizontal

Example #5: Bubba stands at the edge of a building and throws an opossum horizontally over the edge with a speed of 6.50 m/s. The opossum lands 24.1 meters horizontally from the base of the building. How tall is the building? How long for the opossum to land? With what speed and angle of impact does the opossum land? You do not have to solve the questions in the order given. Start with the x – direction and solve for time. vox = 6.50 m/s voy = 0 m/s yo = h = ? y = 0 Dy = 0 – h Dx = 24.1 m

Now that the time is given, follow the last example to find height and final velocity. Look for an equation with Dy and t and solve.

Solve for each component of the final velocity just before impact. For the x – direction: For the y – direction: The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = +6.50 m/s Use Pythagorean theorem: q vy = – 36.3 m/s v Use inverse tangent: below horizontal

Example #6: Bubba stands at the edge of a building and throws an armadillo horizontally over the edge with a speed of 10.0 m/s. The armadillo lands on the ground with a final velocity directed at 55.0o below the horizontal. What is the height of the building? Think of what is given to solve….. vox = 10.0 m/s voy = 0 m/s yo = h = ? y = 0 Dy = 0 – h vy = ?

Next use the vy to solve for the height:

Projectile Motion Day 2 & 3 Projectile HW #2

Projectiles launched at an angle to the horizontal: Yesterday’s notes involved projectiles launched horizontally. In this case the projectiles had a horizontal component to the velocity but not a vertical component. Today the initial velocity will have a magnitude, vo, and an initial angle, qo. The components to the initial velocity are as follows: Angle above the horizontal….. The x – component is adjacent, so use cosine. v vy q vx so The y – component is opposite, so use sine. so

Angle below the horizontal….. vx The x – component is adjacent, so use cosine. q vy v so The y – component is opposite, so use sine. This component also points downwards, so it is negative. so

Example #7: A ball is launched at 25. 0 m/s at an initial angle of 36 Example #7: A ball is launched at 25.0 m/s at an initial angle of 36.9o above the horizontal. (a) What are the x and y components of the initial velocity? vo voy qo vox

(b) What is the greatest height reached by the ball? + voy = +15.0 m/s vy,top = 0 m/s Dy = ? - Recall: Vo = 25.0 m/s

Note: This greatest height can be written as: At what angle should the object be thrown to reach the greatest height? Reason: The largest value that sin(q) can have is +1, and that occurs only when the angle is 90o.

(c) How long did it take the ball to reach the highest point? voy = +15.0 m/s vy,top = 0 m/s t = ?

(d) What was the total time the ball was in the air? voy = +15.0 m/s yo = 0 m yf = 0 m Dy = 0 m t = ? Think of the ball being thrown up vertically with Voy only.

Note: The total time in the air is just twice the time to the top Note: The total time in the air is just twice the time to the top. This is the same result as in chapter 2. There is the same symmetry here as in the purely vertical motion in Ch. 2: When an object starts and stops at the same vertical height, the time to travel to the top is equal to the time to fall back down.

(e) What is the final velocity of the ball? Start with the x – component: For the y – direction: The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = + vocosqo Use Pythagorean theorem: q vy = – vosinqo v Use inverse tangent: below horizontal

(f) What is the range of the ball? The range of a projectile is the horizontal distance the particle travels. (g) Write a general formula for the range using only vo, qo, and g.

Start substituting values in:

Note that 45o gives the greatest range. From your Trig class, you learned (or will learn) of an identity: Note that 45o gives the greatest range.

Example #8: A football is kicked upwards at 75. 0 ft/s at 70 Example #8: A football is kicked upwards at 75.0 ft/s at 70.0o above the horizontal from the top of a 90.0 foot tall building. (a) What is the maximum height of the ball above the ground?

Add this onto the 90.0 ft tall starting height: (b) How long does it take for the ball to hit the ground?

(c) What is the horizontal range of the football?

Stop Tuesday. Do Problems 20, 22, 29 independently. For problem 29 see page 62 for a useful formula.

Warm Up Example #9. 10 Minutes.

Example #9: A cannon fires a round at an angle of 65 Example #9: A cannon fires a round at an angle of 65.0°, and it is in the air for 12.60 s. Find (a) the initial velocity of the projectile.

(b) What is the range of the projectile?

(c) What is the velocity of the projectile, as magnitude and direction, at 10.00 s?

vx = +28.79 m/s Use Pythagorean theorem: q vy = – 36.26 m/s v Use inverse tangent: below horizontal

Example #10: In making a record jump, the truck “Bigfoot” jumped 208 feet. If the truck left the ramp at 69.3 mph, and the landing ramp was identical in angle and height, determine the angle of the launch ramp. Last 4 words spoken by a Redneck? “Hey Y’all, Watch This!”

From example #7, part (g), use:

Example #11: It is the last play of the game and Troy is losing by 2 points to Sunny (cough) Hills. They decide to attempt a 55.0 yard field goal. The kicker kicks the ball straight at the 10.0 foot tall goal posts. If he kicks the ball at 52.5 mph and 40.0° above the horizontal, does Troy win? First, determine the time for the ball to travel 55.0 yard = 165 feet. Then determine the height of the ball (hopefully) above the ground at this time. Is this height greater than 10.0 feet?

Alternatively….. See the solution to problem 29 in the textbook.

Example #12: Jürgen releases a shot-put 2 Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it? Why can’t we use the range formula to find Vo? But…. We could use formula that show the relationships between Y, X, Vo, g, and θ.

Example #12: Jürgen releases a shot-put 2 Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it? Substitute Dx into Dy and solve for vo.

Warm UP to Test Tomorrow. From memory, notes, text or homework write down all useful formulas Homework Check Sample Test Problems 20, 22,29,30,31,32,35, 67, 75