Projectile Motion

Thank you Physics Classroom: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html

What keeps the arrow moving forward?

15 mph

What keeps the arrow moving forward?

• Inertia

•a property of matter that opposes any change in its state of motion

• Newton’s First Law

Projectile

• An object propelled through the air, especially one thrown as a weapon

Why does the arrow fall?

•Acceleration due to gravity

•Newton’s Second Law

•All projectiles are under the influence of gravity

Types of Projectile Motion

.

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html

Projectile Motion

• The process of movement horizontally and vertically simultaneously.

Two Components of Projectile Motion

• Horizontal Motion

• Vertical Motion

• THEY ARE INDEPENDENT OF ONE ANOTHER!!!!!!!!

• The motion of a projectile shows

a typical Parabolic Trajectory or

path

EX 1 A cannon ball is shot from a cannon with a horizontal velocity of 20m/s. How does the vertical and horizontal displacement change as time increases?

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/us

EX 1 A cannon ball is shot from a cannon with a horizontal velocity of 20m/s. What can you say about vertical velocity? Vertical Accelertion?

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/us

EX 1 A cannon ball is shot from a cannon with a horizontal velocity of 20m/s. What can you say about horizontal velocity? Horizontal Acceleration?

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/us

Projectiles Facts1. Projectiles maintain a constant horizontal velocity (neglecting air

resistance) due to 1st law of motion Vi and Vf are equal. We will refer to these as VX (horizontal velocity)

2. Projectiles always experience a constant vertical acceleration of “g” or 9.80 m/s2 (neglecting air resistance) due to 2nd law of motion

3. Horizontal and vertical motion are completely INDEPENDENT of each other.

T Charts and dvvat

•The big formula: d = vit + .5 at2

• dY = viY t + ½ aY t2

• dX = viX t + ½ aX t2

X

Y

To calculate vertical displacementONLY USE VERTICAL INFO !

dy = viy t + ½ ay t2

What is viy t = to?

dy = ½ ay t2

Where:dy = vertical displacement (y axis)ay= g = gravity (9.8m/s2) (some texts use negative to indicate downward. We will assume gravity to be positive.)

t = time in seconds

To calculate horizontal displacement. ONLY USE HORIZONTAL INFO ! Time determined vertically.

dx = vi t + ½ a t2

Since a is zero, then ½ a t2 = zero

dx = vix * t

d = vt

Where:

dx = horizontal displacement (x axis)

The subscript x refers to horizontal

Vix = initial horizontal velocity

t = time in seconds

EX 2

Horizontal (X) Vertical (Y)

d = d =

vi= vi=

vf= vf=

a = a =

t = t =

Make your T chart and enter all your knowns

Horizontal (X) Vertical (Y)

d = 20 m d = 5 m

vi= vi= 0 m/s

vf= vf=

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

t = t =

= vx

Make your T chart and enter all your knowns

d = .5 at2

1.01 s = t

d = vt

d / t = v

20 m / 1.01 s = v

19.8 m/s = v

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high.

A. How long is this ball in flight?

B. How far does the ball travel vertically?

C. How far does the ball travel horizontally?

D. What is the final horizontal velocity?

E. What is the final vertical velocity?

F. What is the overall velocity of the ball as it hits the ground?

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high.

A. How long is this ball in flight? Calculate

B. How far does the ball travel vertically? Given

C. How far does the ball travel horizontally? Calculate

D. What is the final horizontal velocity? Given vf = vi

E. What is the final vertical velocity? Calculate

F. What is the overall (RESULTANT) velocity of the ball as it hits the ground? Calculate

Horizontal (X) Vertical (Y)

d ? Use d = vit + .5at2 d = 15 m

vi= 25 m/s vi= 0 m/s

vf= 25 m/s vf= Use vf = vi + at

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

t = determine from vertical

information

t ? Use d = vit + .5at2

= vx

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high. A. Time in Air:

d = vit + .5at2 Since vi= 0, this can be simplified to:

d = .5at2

To solve for t:

t = d/.5a

1.75 sec

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high. C. Horizontal distance?Use time from vertical motion to calculate distance for horizontal motion

dx = vi t + ½ a t2

Since a is zero, then ½ a t2 = zero

dx = vix * t

d = vt

43.8m

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high. E. Final Vertical Velocity

vf = vi + a t

vf = (9.8m/s2)(1.75s)

vf = 17.15 m/s

Horizontal (X) Vertical (Y)

d = = 43.8 m d = 15 m

vi= 25 m/s vi= 0 m/s

vf= 25 m/s vf= 17.15 m/s

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

t = 1.75 sec t = 1.75 sec

= vx

EX 3. A ball is thrown horizontally at 25 m/s off a roof 15 m high. F. Overall (Resultant) Velocity

vR = (25 m/s)2 + (17.15 m/s)2

Θ = t-1(17.15m/s/25m/s)

vR = 30.32 m/s at 34.45° to the horizontal

25m/s

17.15m/s

VR

• 2 Objects are dropped from a height of 10 m. Object A has a mass of 50 g. Object B has a mass of 100g. If there is no air friction, then:

• A. Object A should hit the ground before Object B

• B. Object B should hit the ground before Object A

• C. Object A and Object B should hit the ground at the same time.

Practice Ex 4 and 5 on your own

EX 4 A projectile is shot with a speed of 39.5 m/s straight off a roof and lands 198 m away. From what elevation was it shot? ? With what speed does it impact the ground vertically and horizontally? With what overall velocity does it impact the ground?

d = ½ at 2

d = vt

d / v = t

198 m / 39.5 m/s = t

5.01 s = t

d = ½(9.8 m/s2)(5.01)

2

d = 123 m

EX 4 A projectile is shot with a speed of 39.5 m/s straight off a roof and lands 198 m away. From what elevation was it shot? ? With what speed does it impact the ground vertically and horizontally? With what overall velocity does it impact the ground?

vyf

= at

vxf

= 39.5 m/s

vyf

= (9.8 m/s2)(5.01)

vyf

= 49.1 m/s

vr = √(49.1 m/s)

2+ (39.5 m/s)

2

vr= 63.0 m/s

EX 5 A projectile is shot horizontally off a 267-m tall building with a speed of 14.3 m/s.

A. With what speed does it impact the ground

vertically and horizontally?

B. With what overall velocity does it impact the

ground?

Vertical (Y) Horizontal (X)

d = 267 m d = Use d = vt

vi= 0 m/s vi= 14.3 m/s

vf= Use vf = vi + at or

vf2 = vi

2 + 2ax

vf= 14.3 m/s

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

t = Use d = vit + .5at2 t = determine from vertical

information

= vx

vf horizontal is constant at 14.3 m/s

• vf2 = vi

2 + 2ax to determine vf vertically

• vfy = 72.3 m/s

• overall velocity?

• This is just determining the resultant using Pythagoreans

• vr2 = (14.3 m/s)2 + (72.3m/s)2

• vr = 73.7 m/s

• Supposing a snowmobile is equipped with a flare launcher which is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)?

• a. in front of the snowmobile

• b. behind the snowmobile

• c. in the snowmobile

Flipping Physics Frame of Reference and Projectile Motion

https://www.youtube.com/watch?v=mYH_nODWkqk

•Many would insist that there is a horizontal force acting upon the ball since it has a horizontal motion. This is simply not the case. The horizontal motion of the ball is the result of its own inertia. When projected from the truck, the ball already possessed a horizontal motion, and thus will maintain this state of horizontal motion unless acted upon by a horizontal force. An object in motion will continue in motion with the same speed and in the same direction ... (Newton's first law). Remind yourself continuously: forces do notcause motion; rather, forces cause accelerations

Where will the package land? (neglect air resistance )

Ex. 6 A plane flying at 115 m/s drops a package from

600m. How far from the drop point will it land?

Objects dropped from a moving vehicle have the same

horizontal velocity as the moving vehicle.

Horizontal (X) Vertical (Y)

d = d = 600 m

vi= 115 m/s vi=

vf=

115 m/s

vf=

a = a =

t = t =

= vx

Horizontal (X) Vertical (Y)

d = 1273.05 d = 600 m

vi= 115 m/s vi= 0 m/s

vf=

115 m/s

vf=

a = 0 a = 9.9 m/s2

t = 11.07 s t = 11.07 s

= vx

Horizontal:

Vx = 115 m/s

dx = ?

Vertical:

Viy = 0

dy = 600. m

a = 9.8 m/s2

This is the same problem we’ve been working…

dy = ½ at2

600. m= ½ (9.8m/s2)t2

t = 11.1 s

dx = (115 m/s)(11.1s)

dx = 1273.05 m

Example 7: A projectile is thrown upward at a rate of 13.22 m/s and at an angle of 83.1° with the horizontal.

A. How long is the projectile in the air?

B. Calculate the range.

C. What is the peak height?

What can you say about a trajectory path?

Example 7: A projectile is thrown upward at a rate of 13.22 m/s and at an angle of 83.1°with the horizontal.Indicate knowns

Horiz

(X)

Vert Up

(Y)

Vert down

(Y)

d

vi 0 m/s

vf 0 m/s

a 0 m/s2 9.8 m/s2 9.8 m/s2

t

=vx

How do we determine the initial velocities?

Given 13.22 m/s at an angle of 83.1°

This describes the resultant of the horizontal and vertical velocity components.

You need to determine the horizontal and vertical components

Vertical

Sin (83.1°) (13.22 m/s)

83.1°

Horizontal

Cos (83.1°) (13.22 m/s)

Example 7: A projectile is thrown upward at a rate of 13.22 m/s and at an angle of 83.1°with the horizontal.Indicate knowns

Horiz

(X)

Vert Up

(Y)

Vert down

(Y)

d

vi 1.59 m/s -13.1 m/s 0 m/s

vf 1.59 m/s 0 m/s 13.1 m/s

a 0 m/s2 9.8 m/s2 9.8 m/s2

t

= vx

Time at Peakt = vfy - viy

ay

13.1 m/s – 0 m/s

9.8m/s2

t = 1.34 s

Horizontal Time would be 2.68 sec

Time

Peak Height

d = .5at2

(.5)(9.8 m/s2)(1.34 s)2

8.80m

Horizontal Displacement(Remember to double time)

• dx = vix•t

• dx = (1.59 m/s)(2.68 s)

• dx = 4.26 m

Projectiles at a known velocity and angleSteps to determine time, height , and range

1. Determine X component (C=A/H)

This yields the horizontal vi and vf

2. Determine Y component (S=O/H)

This yields the vertical up vi and vertical down vf

3. Make 3 column table of knowns: Horizontal, Vertical Up, and Vertical down

Remember horizontal acceleration = 0; vertical acceleration is 9.8 m/s2 due to gravity

4. Calculate peak time using vertical down column vf = vi + at

5. Total time in air (horizontal) is 2 x peak time

6. Calculate peak height using vertical information x = .5at2

(vi t = 0 in vertical down column)

7. Calculate range using horizontal information x = vi t (.5at2 = 0)