+= ? The wrong diagrams Draw the right diagram for A + B do now A B.

+ = ?

• The wrong diagrams

Draw the right diagram for A + B

do now

AB

3.3 projectile motion

Objectives

1. Recognize examples of projectile motion.

2. Describe the path of a projectile as a parabola.

3. Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion.

question• The long jumper builds up speed in the x-

direction and jumps, so there is also a component of speed in the y direction. Does the angle of take-off matter to the jumper?

• A parabola is the set of all points in the plane equidistant from a given line (the conic section directrix) and a given point not on the line (the focus).

y ~ x2

What is a projectile?

• An object that is launched into the air with some INITIAL VELOCITY

• Can be launched at ANY ANGLE

• In FREEFALL after launch (no outside forces except force of gravity)

• The path of the projectile is a PARABOLA

Effect of air resistance

demo

• The vertical motion of a projectile is same as free falling object, with constant acceleration:

• a = -9.81 m/s/s

Two types of projectiles

• Projectile launched horizontally

• Projectile launched at an angle

Projectile Launched horizontally

Projectile Launched at an angle

Over the EdgeHorizontal Projectiles

Projectiles Launched Horizontally

• Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.

Vertical motion:• Free fall free rest:

vy = 0 ay = -9.81 m/s2

Horizontal motion:• vx = vix = constant • ax = 0

Velocity of Horizontal Projectiles• Horizontal motion is

constant: velocity is constant.

• Vertical: same as drop the ball from rest: velocity is increasing by 9.81 m/s every second

Projectile launched horizontally• A projectile is any object upon which the only force is

gravity• Projectiles travel with a parabolic trajectory due to the

influence of gravity, • In horizontal direction, the projectile has no

acceleration, its velocity is constant: vx = vi

• In vertical direction, the projectile has acceleration: a = -9.81 m/s/s. Its initial vertical velocity is zero and it changes by -9.81 m/s each second. Same as an object falling from rest.

• The horizontal motion of a projectile is independent of its vertical motion

Example• An object was projected horizontally from a tall cliff.

The diagram represents the path of the object, neglecting friction.

• Comparing the following at point A & B:1. Acceleration2. Horizontal velocity3. Vertical velocity

As the red ball rolls off the edge, a greenball is dropped from rest from the same

height at the same time

Which one will hit the ground first?

They will hitat the SAME

TIME!!!

The same time?!? How?!?

vix

The red ball has an initialHORIZONTAL velocity (vix)

But does not have any initialVERTICAL velocity (viy = 0)

The green ball falls from restand has no initial

velocity IN EITHERDIRECTION!

viy and vix = 0

One Dimension at a Time• Both balls begin with no VERTICAL

VELOCITY

• Both fall the same DISTANCE VERTICALLY

• Find time of flight by solving in the appropriate dimensionWe can find an object’s displacement in

EITHER DIMENSION using TIME

Example #1

• A bullet is fired horizontally from a gun that is 1.7 meters above the ground with a velocity of 55 meters per second.

• At the same time that the bullet is fired, the shooter drops an identical bullet from the same height.

– Which bullet hits the ground first? Both hit the ground at the same time

Equation for horizontal projectile

Horizontal• ax = 0

Vertical • ay = -9.81 m/s2

d = ½ (vi + vf)t

vf = vi + at

d = vit + ½at2

vf2 = vi

2 + 2ad

vix = vicosθ = vi

x = vix∙t

viy = visinθ = 0

y = ½ (viy + vfy)t

vfy = viy + ayt

y = viyt + ½ayt2

vfy2 = viy

2 + 2ayy

x and y has the same t

θ = 0

Example #2• An airplane making a supply drop to

troops behind enemy lines is flying with a speed of 300 meters per second at an altitude of 300 meters.

– How far from the drop zone should the aircraft drop the supplies?

Need time from vertical

dy = viyt + ½ ayt2

300 m = 0 + ½ (-9.81 m/s2)t2

t = 7.82 s

Use time in horizontal

dx = vixt + ½ axt2

dx = (300 m/s)(7.82 s) + 0dx = 2346 m

Example #3• A stuntman jumps off

the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building.

– What minimum initial velocity does he need in order to make it onto the air mattress?

Need time from vertical

dy = viyt + ½ ayt2

45 m = 0 + ½ (-9.81 m/s2)t2

t = 3.03 s

Use time to find v

v = d / tv = 15 m / 3.03 s

v = 4.95 m /s

Example #4• A CSI detective investigating an accident

scene finds a car that has flown off the edge of a cliff. The car is 79 meters from the edge of the 25 meter high cliff.

– What was the car’s initial horizontal velocity as it went off the edge?

Need time from vertical

dy = viyt + ½ ayt2

25 m = 0 + ½ (-9.81 m/s2)t2

t = 2.26 s

Use time in horizontal

dx = vixt + ½ axt2

79 m = vix (2.26 s) + 0vix = 34.96 m/s

Example #4

Example #5• The path of a stunt car driven horizontally off a cliff is

represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second. 1. Determine the magnitude of the horizontal

component of the velocity of the car at point B. [Neglect friction.]

2. Determine the magnitude of the vertical velocity of the car at point A.

3. Calculate the magnitude of the vertical displacement, dy, of the car from point A to point B. [Neglect friction.]

Class work• Page 101 - Sample problem 3D

• Page 102 – practice 3D

Answers:

1.0.66 m/s

2.4.9 m/s

3.7.6 m/s

4.5.6 m

Do now

Fire Away!!!Projectiles

Launched at an Angle

Projectile Launched at an angle

Projectile Vector Diagram

t1/2 vfy = 0 (at top)

ttot

Initial velocity

30°

What is the horizontal part ofthe soccer ball’s initial

velocity?vix = vi cos θ

What is the vertical part ofthe soccer ball’s initial

velocity?viy = vi sin θ

12 m/s

6 m/s

10.4 m/s

Pythagorean Theoremvi

2 = vix2 + viy

2

What do we know or assume about the vertical part of a projectile problem?

• Initial vertical velocity = vi sin θ

• Acceleration = -9.81 m/s2

• Vertical speed will be 0 at the maximum height

• Time to top = HALF total time in the air

– Find time to top using final velocity equation

• Vertical Distance – Max height

– Use time to top and solve vertical distance equation

What do we know or assume about the horizontal part of a projectile problem?

• Initial vertical velocity = vi cos θ

• Acceleration = 0 (if we assume no air resistance)

• Horizontal Distance – Range– Use total time and solve horizontal distance equation

The symmetrical nature of a ground launched projectile

•What is the acceleration at the top of the path?•What is the vertical velocity at the top of the path?

Projectile launched at an angle

• A projectile is any object upon which the only force is gravity

• Projectiles travel with a parabolic trajectory due to the influence of gravity,

• In horizontal direction, the projectile has no acceleration, its velocity is constant: vx = vicosθ

• In vertical direction, the projectile has acceleration: a = -9.81 m/s/s. Its velocity of a projectile changes by -9.81 m/s each second. Same as a free falling object.

• The horizontal motion of a projectile is independent of its vertical motion

• A projectile is fired with initial horizontal velocity at 10.00 m/s, and vertical velocity at +20. m/s. Determine the horizontal and vertical velocity at 1 – 5 seconds after the projectile is fired. Use g = 10 m/s/s.

TimeHorizontalVelocity

VerticalVelocity

Accelartion

0 s

1 s

2 s

3 s

4 s

5 s

Equation for projectile motion

Horizontal• ax = 0

Vertical • ay = -9.81 m/s2

1D equation:

d = ½ (vi + vf)t

vf = vi + at

d = vit + ½at2

vf2 = vi

2 + 2ad

vix = vicosθ

x = vix∙t

viy = visinθ

vfy = 0 (at top)

vfy = - viy (at same height)

y = ½ (viy + vfy)t

vfy = viy + ayt

y = viyt + ½ayt2

vfy2 = viy

2 + 2ayy

• Since velocity is a vector quantity, vector resolution is used to determine the components of velocity.

θ

vi

vix

viy

SOH CAH TOA

sinθ = viy / vi

viy = visinθ

cosθ = vix / vi

vix = vicosθ

• Special case: horizontally launched projectile:

• θ = 0o: viy = visinθ = 0; vix = vicosθ = vi

Initial Velocity Components

vi2 = vix

2 + viy2

1. A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.

2. A motorcycle stunt person traveling 70 mi/hr jumps off a ramp parallel to the horizontal.

3. A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.

Examples Determine the horizontal and vertical components

example• A machine fired several projectiles

at the same angle, θ, above the horizontal. Each projectile was fired with a different initial velocity, vi. The graph below

represents the relationship between the magnitude of the initial vertical velocity, viy, and the magnitude of

the corresponding initial velocity, vi, of these projectiles. Calculate

the magnitude of the initial horizontal velocity of the projectile, vix, when the magnitude

of its initial velocity, vi, was 40.

meters per second.

• The point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters as – the horizontal displacement, – the vertical displacement, – the final vertical velocity, – the time to reach the peak of the trajectory, – the time to fall to the ground, etc.

Example• A football is kicked with an initial velocity of 25 m/s at an

angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

Horizontal Component Vertical Component

ax = 0 m/s/s

vix = 25 m/s•cos45o= 17.7 m/sx = ?

ay = -9.81 m/s/s

viy=25 m/s•sin45o= 17.7 m/s

vfy = -17.7 m/s

ymax = ?

ttotal = ?

• Solve for t – use vertical information:

t = 3.61 s

• Solve for x – use horizontal information:

x = vix•t + ½ •ax•t2

x = 63.8 m

vfy = viy + ayt

• Solve for peak height – use vertical information:

vfy2 = viy

2 + 2ayypeak

At the top, vfy = 0

ypeak = 15.9 m

example• A long jumper leaves the ground with an initial velocity of

12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

Horizontal Component Vertical Component

ax = 0 m/s/s

vix = vi•cosθ

vix = 12 m/s•cos28o

vix = 10.6 m/s

ttotal = 2tup x = ?

ay = -9.8 m/s/s

viy = vi•sinθ

viy = 12 m/s•sin28o

viy = 5.6 m/s

vfy = -5.6 m/sy = ?

tup = ?t = 1.1 s x = 12.2 m ypeak = 1.6 m

Example • A cannon elevated at an angle of 35° to the

horizontal fires a cannonball, which travels the path shown in the diagram.  [Neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.] If the ball lands 7.0 × 102 meters from the cannon 7.0 seconds after it was fired,

1. what is the horizontal component of its initial velocity?

2. what is the vertical component of its initial velocity?

3. An object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second.  The object strikes the ground 3.0 seconds later.  What is the vertical speed of the object as it reaches the ground? [Neglect friction.]

4. In the diagram, a 10.-kilogram sphere, A, is projected horizontally with a velocity of 30. meters per second due east from a height of 20. meters above level ground. At the same instant, a 20.-kilogram sphere, B , is projected horizontally with a velocity of 10. meters per second due west from a height of 80. meters above level ground. [Neglect air friction.] Initially, the spheres are separated by a horizontal distance of 100. meters. What is the horizontal separation of the spheres at the end of 1.5 seconds?

Total flight time, range, max height Time to go up

• t = visinθ / g

• As θ increases, flight time increase. Max time: θ = 90o

Max height

• hmax = (visinθ)2/2g

• As θ increases, flight height increase. Max height: θ = 90o

Range

• Range = vi2sin2θ /g

• Projectile has maximum range when θ = 45o

Class work

• Page 103 – sample problem 3E• Page 104 – practice 3E #1-5

Answers:

1.Yes

2.70.3 m

3.20. s; 4.8 m

4.6.2 m/s

5.17.7 m/s; 6.60 m