PROJECTILES Wile E Coyote Effect I’m sure all of you have seen the cartoon where the Coyote is...

PROJECTILES

Wile E Coyote Effect

• I’m sure all of you have seen the cartoon where the Coyote is chasing after the Road Runner and runs off of the cliff. He hangs in mid air for a second, looks down, and then starts to fall.

• The question is how true is this, and how many people believe it is true?


A few years ago some researchers in the U.S. went to elementary schools, junior and senior high schools, and universities and asked them to look at the following:

"Ignoring air resistance, which of the following correctly shows what an object would do if it rolled off a cliff?"


1. The object will stop in midair, and then start to fall straight down.

2. The object will move forward at first, but will eventually just fall straight down.

3. The object will continue to move forwards the entire time it is falling.


1. About 60% said number 1 was correct. Because this is what the Coyote always did in cartoons (some people actually referred to it in their explanation of why they chose this

answer), the researchers called it the Wile E. Coyote Effect.

2. About 25% said number 2 was correct.

3. Only about 15% answered number 3.


C is the correct answer. The object will continue to move forwards the entire time it is falling.

Background

Assumptions for the motion of the object:

All forces acting on it are balanced

Fnet = 0 State of ‘equilibrium’

Newton’s first law of motion, (the law of inertia) – An object at rest remains at rest until an

unbalanced force acts upon it.– An object in motion continues in motion (straight

line, constant speed) until an unbalanced force changes that motion.

Background

Horizontal and Vertical components of a vector

Kinematics is not limited to 1DIn two dimensions, there are the x and y axesValues are split into two different components

(x and y components)Movement in the x axis does not affect movement in the y axis and vice versa

VECTOR RESOLUTION

Trigonometry:– Use sine to determine the length of the side

opposite the angle. – Use cosine to determine the adjacent side.

46.0

46.0 N

Calculate the resultant’s magnitude by:

and direction by:

• The components of a vector may (must!) be treated separately!

22yx CCC

x

y

C

C1tan

VECTOR RESOLUTION

Projectile

• A projectile is any object which, when thrown, shot or otherwise projected, continues in motion by its own inertia and is influenced only by the downward force of gravity.

Note: A force is not required

to keep an object in motion!

Types of Projectiles

1. An object dropped from rest (1D)2. An object which is thrown vertically

upwards (1D)3. An object which is thrown upwards at an

angle (2D)• Horizontally launched• Launched at an angle

In all cases, air resistance is assumed to be negligible.

Trajectory

The path of any projectile is called a trajectory.

Objects Launched Horizontally

• If there were NO GRAVITY, the canon ball would move at constant speed in a straight line.

• On earth there IS gravity, so the ball would fall towards earth as it moves horizontally

• Indicates two types of motion

– horizontal constant velocity

– vertical – constant acceleration

Gravity

Gravity affects what component?

Gravity acts only in a downward direction, and makes objects fall toward earth in the downward direction at a rate of 9.8 m/s2.

• So gravity ONLY affects the Vertical Motion on an object

• The horizontal motion of an object IS NOT affected by gravity!

A Dropped Question

If two objects are released simultaneously from the same height one object is launched horizontally and the other is dropped vertically which one will hit the ground first?

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A Dropped Question

BOTH will hit the ground at the same time! • Both objects are being accelerated

towards earth by the same force - gravity!

Cannon Example

Is there any horizontal force acting on the ball once it leaves the canon?

• The horizontal component of the ball’s motion is constant.

• This is true for all projectiles on earth.

A cannon is fired horizontally off a high cliff.

Cannon Example

Is there any vertical force acting on the canon ball once it leaves the canon?

• The vertical component of the ball’s motion is accelerating downward.

• This is also true for all projectiles on earth.

A cannon is fired horizontally off a high cliff.

The SCOOP

THE HORIZONTAL COMPONENT OF MOTION FOR A PROJECTILE IS COMPLETELY

INDEPENDENT OF THE VERTICAL COMPONENT OF MOTION.

EACH ACTS INDEPENDENTLY OF THE

OTHER.

THEIR COMBINED EFFECTS PRODUCE THE PATH THAT IS FOLLOWED BY THE

PROJECTILE.

Onto the Numbers

• A projectile shot horizontally:

Notice:

1. The horizontal component (vx) of the velocity is the same everywhere!

2. The vertical component (vy) of the velocity is the same as if it were simply falling!

Key Terms for Projectiles

• Projectile: the object being launched• Trajectory: the path followed by the projectile• Range: the horizontal distance traveled by the

projectile• vx means: horizontal velocity• vyi means: initial vertical velocity• vyf means: final vertical velocity• dx means: horizontal distance (aka - range)• dy means: vertical distance (aka - height)

Horizontally Launched Projectiles

Horizontal Component• vx is a constantfor displacement (s) or

distance (d)• dx = vxt

Vertical Component• vy = -gt

for displacement (s) or distance (d)

• dy = - ½ gt2

Where g = 9.8 m/s2Proof:for displacement d

if dx = vxt and dy = - ½ gt2

then t = dx substituting dy = - ½ g (dx/ vx )2

vx -½ g and vx are constants, rearranging we getdy = (-g / 2vx

2) dx2 (-g / 2vx

2) = constant = kTherefore,

y = k x2

Recognize the formula for a parabola ?


A cannon is fired off a high cliff with a horizontal speed of 100m/s. If it takes 12 seconds for the ball to hit the ground, how high is the cliff and how far did the cannon ball travel from the base of the cliff?

Step 1: Draw a picture.

Step 2: Break the motion into horizontal and vertical components. Identify the knowns. the unknowns, the correct equation and solve. (Solve the acceleration problems.)


A cannon is fired off a high cliff with a horizontal speed of 100m/s. If it takes 12 seconds for the ball to hit the ground, how high is the cliff and how far did the cannon ball travel from the base of the cliff?

Horizontal Motion Vertical Motion

vx=100 m/s viy=0 m/s

t = 12 s t = 12 s

a = 0 m/s2 a = -9.8m/s2

dx= ? dy= ?

dx = vxt = dy= viyt+1/2at2

dx = 1200 m

dy= -705.6 m

100 (12) =

dy= ½(-9.8)(12)2

Airplane Drop

If a package is dropped from an airplane flying with a horizontal speed of 115 m/s at an altitude of 1050 m where will the package land?

a. Behind the plane

b. Under the plane

c. In front of the plane

Airplane Drop

Hippo Drop

From a height of 784 m a hippo falls from an airplane flying horizontally at 30 m/s. Oh no! His chute is too small and he falls like he isn’t even wearing one. What horizontal distance does he travel before striking the ground?

Horizontal Motion Vertical Motionvx= viy=t = t =

a = dx= dy=

dy=viyt+1/2at2

dx = vxt =

dx = 379.5 mt = 12.65 s

30 m/s 0 m/s? ?

-9.8 m/s2

-784 m?

a

dt

2

)8.9(

)784(2

30(12.65)

Bullseye or Not

• An arrow is fired directly at a bull’s eye of a target that is 60 m away. The arrow has a horizontal speed of 89 m/s. When the arrow is fired, it is exactly 1 m above the ground. It was a lousy shot and a miss. How far short of the target was it?

Challenge Shot

• From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 343 m/s, parallel to the ground. The bullet puts a hole in a window of another window and hit the wall that faces the window. Determine the distances D and H from the picture. Assume the bullet does not slow down as it passes through the window.

Daredevil

A motorcycle is traveling horizontally at take-off. In flight it follows a projectile path, so we can treat its vertical and horizontal motions separately. To reach the landing-zone the bike must travel 48 m horizontally in the same time as it travels 19.62 m vertically. What speed must it be traveling at initially to land in the center of the landing zone.

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Vertically Launched Projectile

If a ball is launched vertically off the back of a moving pickup truck where will it land?

• A. behind the truck• B. in the bed of the truck• C. in front of the truck.

Projectiles

• Projectiles follow a parabolic path because of the force of gravity

• According to Newton’s 1st law: An object in motion will stay in motion unless acted upon by an outside force

• However, gravity acts upon the projectile and causes the velocity of the y axis to become negative after arriving at the maximum height, resulting in the projectile moving downwards

• When the projectile hits the ground, friction is what causes the halt of the object in terms of the x axis

• What determines the displacement of the x value is the time that the projectile is in the air

Projectiles Launched at an Angle

Types• A projectile launched off a cliff at an angle

and then falls to some point below the cliff

• A projectile launched from ground level and return to ground level.


• If a cannonball shot at an upward angle with no gravity, the ball would not fall toward earth, but would rather follow a straight line shown by the dashed line.

• With gravity the ball continually falls beneath the imaginary dashed line until it strikes the ground.


• Initial vertical velocity is not zero. (viy ≠ 0)• Given only the resultant velocity of the

projectile and the angle at which it was launched.

• This resultant velocity must be broken into its horizontal and vertical components.

Velocity Components

• Horizontal velocity (vx): vx = vi cos θ

• Vertical velocity (viy): viy = vi sin θ

Copyright© 2007 Pearson Prentice Hall, Inc.


Notice:1. The horizontal

component (vx) of the velocity is the same everywhere!

2. The vertical component (vy) of the velocity is the same as if it had been thrown upwards with vi = 20 m/s!


• As it rises, its velocity decreases with time due to gravity. At its highest point, its vy = 0.

• As it falls from its highest point vy increases as it travels with gravity.

• Its vertical velocity is just like an object that is thrown straight up.

Components

Remember, for vertical components the constant acceleration equations can be used.

For horizontal components we use the constant velocity equation.

vf = vi + at

d = ½ (vf + vi)t

d = vit + ½ at2

vf2 = vi

2 + 2ad

vx =dx/t


Note: Principle of independent velocities still holds

REMEMBER!!!For dy and vy downward is negativeVi can be resolved into the components: vx

and vy

Vy = Visinθ Vy decreases due to gravity

Vx = Vicosθ Vx remains constant

The total displacement of the projectile is called the range

60°

viy

vix

Projectile Problem

A cannonball is shot upwards at a 60° angle to the ground and has an initial velocity of 200 m/s.

(a) What are the horizontal and vertical components of its velocity?

(b) How long was the cannonball in the air?

(c) How far did the cannonball go?

(d) How high was the cannon ball after 4 seconds had passed?

Projectile Problem


(a) What are the horizontal and vertical components of its velocity?

vi = 200 m/s Horizontal Vertical

vx = vicosθ viy = visinθ

vx = 200cos(60) viy = 200sin(60)

vx = 100 m/s viy = 173 m/s

60°

voy

vx

V o =

200

m/s


(b) How long was the cannonball in the air? Horizontal Vertical vx = 100m/s viy = 173 m/s

t = ? t = ? a = -9.8 m/s2

vfy = 0 m/s at top of path

vfy = vyi + at

t = 17.65 s (time to top of path)

t = 2(17.65) = 35.3 s

Projectile Problem

60°

viy

vx

V o =

200

m/s

a

vvt oyfy

8.9

1730


(c) How far did the cannonball go? Horizontal Vertical vxi = 100m/s vyi = 173 m/s

t = 35.3 s t = 35.3 s

dx = ? a = -9.8 m/s2

dx = vxt

dx = 100 (35.3)

dx = 3530 m

Projectile Problem

60°

viy

vix

V o =

200

m/s


(d) How high was the cannon ball after 4 seconds had passed?

Horizontal Vertical vxi = 100m/s vyi = 173 m/s

t = 4 s a = -9.8 m/s2

dy = ?

dy = vyit+1/2at2

dy = 173 (4) + ½(-9.8)(4)2

dy = 614 m

Projectile Problem

60°

viy

vix

V o =

200

m/s

Projectile Problem

A football punter kicks the ball with an initial velocity of 30m/s at an angle of 45°

a. What is the range of the kick?

b. What is the max height of the kick?

c. What is the hangtime of the kick?

Stuntman

John a motorcross rider decides to demonstrate his skill by jumping school buses. If he sets up a launching and landing ramp at 18° and the height of the school bus and launches himself at an initial velocity of 33.5 m/s how many whole school buses could he clear if each bus is 2.74 m wide?

Slugger

Recently, at Turner Field you caught a Chipper Jones homerun. Always having physics on you mind, you calculated the velocity it hit your glove at 36 m/s and an angle of 28°. Your glove was 7.5 m above the height Chipper hit the ball. What was the initial velocity that Chipper hit the ball (magnitude and direction)?

Angle vs. Range

Which launch angle will result in the greatest range?

A. 30°

B. 45°

C. 60°Which angle went the highest?

Which angle had the most

hang time?

Angle vs. Range

• The diagram shows that the projectiles reach different altitudes or heights above the ground and also have different ranges.

What two angles result in the same range?

• If we add these two angles up they total 90°.

• These are called complimentary angles

White Board Challenge

• A battleship simultaneously fires two shells toward two enemy ships, one close by (A), one far away (B). The shells leave the battleship at different angles and travel along the parabolic trajectories indicate below. Which of the two enemy ships get hit first? Explain.

Monkeying Around

If a monkey hunter was aiming at a monkey in a tree and the monkey lets go the instant the hunter fires. Where should the hunter aim?

A. directly above the monkey

B. directly at the monkey

C. directly below the monkey

Monkeying Around

What if a monkey hunter was now aiming at a monkey in a tree and the monkey lets go the instant the hunter fires but this time with a much slower gun. Where should the hunter aim?

A. directly above the monkey

B. directly at the monkey

C. directly below the monkey

Range

• The horizontal distance a projectile travels before returning to its original height.

• Range can be determined when the angle () and initial velocity (vi) are known:

• “g” is acceleration due to gravity = 9.8 m/s2 (on earth)

• vi is the velocity at the angle (θ) given.

gRvi )2sin(

2

Max Height

• Solving for the height (H) when vfy = 0 results in the following equation:

• Where H =dymax = max height

vi = initial velocity

θ = angle of launch• For the height at any time

dy = vi sinθ t – ½ gt2 or dy = vyt – ½ gt2

g

vdH iy 2

sin 2

max

Total Time of Flight

Solving for the time at the maximum height and then doubling it, yields this relationship for total time (T) of flight:

org

vT i sin2

g

v yi2

Understanding Check

• At the instant a HORIZONTALLY held rifle is fired over ground level, a bullet held at the side of the rifle is released and drops to the ground. Which bullet hits the ground first?

• At the instant a rifle HELD AT AN UPWARD ANGLE is fired over ground, a bullet held at the side of the rifle is released and drops to the ground. Which bullet hits the ground first?

• A projectile is launched at an angle into the air. What is the acceleration of its vertical components of motion? Of its horizontal component?

• At what part of its trajectory does a projectile have minimum speed? EXPLAIN!!!!

Effect of Air Resistance

v0=44.7 m/s

θ = 60°

Path I (Air) Path II (Vacuum)

Range 98.5 m 177 m

Max Height 53.0 m 76.8 m

Time of Flight 6.6 s 7.9 s

Effect of Air Resistance

Key Differences:

Path is no longer parabolic

The maximum height and range are less than without air resistance

The angle at which the projectile impacts the ground is steeper.

PROJECTILES Wile E Coyote Effect I’m sure all of you have seen the cartoon where the Coyote is...

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