Objects launched are projectilesObjects launched are projectiles balls, bullets, arrows, space ships…balls, bullets, arrows, space ships…

The The PATHPATH a projectile follows is the a projectile follows is the TRAJECTORYTRAJECTORY

The trajectory for a projectile is a The trajectory for a projectile is a PARABOLAPARABOLA

We will We will ignore ignore wind wind

resistancresistancee

Motion for a Projectile is Motion for a Projectile is described in terms of…described in terms of…

PositionPosition VelocityVelocity

AccelerationAcceleration All three of these are All three of these are vector quantitiesvector quantities

and will be represented with arrows like and will be represented with arrows like the previous topicthe previous topic

We must remember that We must remember that horizontalhorizontal and and verticalvertical velocities (components) of a velocities (components) of a projectile are projectile are INDEPENDENTINDEPENDENT! !

Drop a ballDrop a ball

AcceleratAccelerates es

DownwarDownwardd

Roll a BallRoll a Ball

Rolls across Rolls across

Constant Velocity (No Constant Velocity (No friction)friction)

Something thrown Something thrown through the air will have through the air will have both a down both a down (vertical)(vertical) and and across across (horizontal)(horizontal) velocity velocity

1.1. accelerate down accelerate down (due to gravity)(due to gravity)

2.2. go across at constant go across at constant velocityvelocity(we assume no air resistance)(we assume no air resistance)

Each velocity will Each velocity will NOTNOT affect the affect the

other other (Independence)(Independence) VVh h

(V(Vxx))

VVvv (V(Vyy))

VVxx is is ConstantConstant

VVyy is is changing changing

(accelerating b/c of (accelerating b/c of gravity)gravity)

What is the time of flight?What is the Vx?

EquationsEquations

PastPast

D = VtD = Vt

D= VD= Viit + ½ att + ½ at22

VVff= V= Vii + at + at

PresentPresent

X = VX = Vxxtt

Y = VY = Vyiyi t+ ½ gt t+ ½ gt22

Y = ½ gtY = ½ gt22

tt22 = = 2y2y//gg

VVyfyf = V = Vyiyi + gt + gt

Sample ProblemSample Problem A stone is thrown horizontally A stone is thrown horizontally

at +15 m/s from the top of a at +15 m/s from the top of a cliff 44m high.cliff 44m high.

1.1. How long does the stone take to How long does the stone take to reach the bottom of the cliff?reach the bottom of the cliff?

2.2. How far from the base of the cliff How far from the base of the cliff does the stone strike the grounddoes the stone strike the ground

3.3. Sketch the trajectory of the stoneSketch the trajectory of the stone

44m44m

15 15 m/sm/s

GivenGiven

VVxx = 15m/s = 15m/s

VVyiyi = 0 m/s = 0 m/s

G= G= -9.8m/s-9.8m/s22

y = -44my = -44mFormulaFormula

X = VX = Vxxtt

Y = VY = Vyyt + ½ t + ½ gtgt22

SOLUTIONSOLUTIONa. ta. t22 = 2 = 2 yy//gg

tt22 = = 2(-44)2(-44)/ / -9.8-9.8

t = t = √√9 t= 3sec9 t= 3sec

b. X= Vb. X= Vxxtt

= (15 = (15 mm//ss) (3s)) (3s)= 45m= 45m

4444mm

t= 3sect= 3sec

45m45m

Practice ProblemsPractice Problems

1.1. A stone is thrown horizontally at a A stone is thrown horizontally at a speed of speed of +5.0+5.0mm//ss from the top of a cliff from the top of a cliff 78.4m high78.4m high..

a.a. How longHow long does it take the stone to does it take the stone to reach the bottom of the cliff?reach the bottom of the cliff?

b.b. How farHow far from the base of the cliff does from the base of the cliff does the stone strike the ground?the stone strike the ground?

2.2. A steel ball rolls with a constant velocity A steel ball rolls with a constant velocity across a across a tabletop .950m hightabletop .950m high. It rolls off . It rolls off and and hits the ground +.352mhits the ground +.352m horizontally horizontally from the edge of the table. from the edge of the table. How fast was How fast was the ball rollingthe ball rolling??

Homework P. 102

1-3

Page 118 K and L

Objects Launched at Objects Launched at AnglesAngles

When we analyze a When we analyze a trajectory, we should see trajectory, we should see what happens to vertical what happens to vertical

and horizontal and horizontal components to components to

understand the motionunderstand the motion

Page 118 J

Find component velocities on the way up.

Find Resultant velocities on the way down.

Shallow AngleShallow Angle

Steep Angle

Use Vector Knowledge to Use Vector Knowledge to Solve for Solve for

VVxx & V & Vyy

The launching The launching angleangle and the and the velocityvelocity determine determine how farhow far the the

object will travel.object will travel. When the launching speed or When the launching speed or

force is the same, the angle alone force is the same, the angle alone will determine the range.will determine the range.

RANGERANGE: how far (horizontally) a : how far (horizontally) a projectile goesprojectile goes

NOTENOTE

For working with angled For working with angled trajectories, the launching height is trajectories, the launching height is

usually the same as the landing usually the same as the landing height When it is height When it is Y = 0!Y = 0!

Finding Time for angled launches

Time = 2Vy

g

Only works when landing/launching heights are the same. I call this the “hangtime” formula.

Example Problem for a projectile shot at Example Problem for a projectile shot at an angle…an angle…

A small metal ball is shot with a A small metal ball is shot with a velocity of 4.47velocity of 4.47mm//ss at an angle of at an angle of 6666°° above the horizontal. above the horizontal.

a.a. How long does it take the ball How long does it take the ball to land?to land?

b.b. How high did the ball fly?How high did the ball fly?

c.c. What was the range?What was the range?

GivenGiven

VVii = = 4.474.47mm//ss

0 = 66°0 = 66°

VVxx

VVyy4.44.477 6666

FormulaFormula

y = Vy = Vyyt + ½ t + ½ gtgt22

X = VX = Vxxtt

SolutionSolution

A.A. when land when land y=0y=0

0 = V0 = Vyyt + ½ gtt + ½ gt22

-V-Vyyt = ½ gtt = ½ gt22

t = -2Vt = -2Vyy/ g/ g

t = t = -2(4.08)-2(4.08)//-9.8-9.8

VVyy = 4.47(sin 66) = = 4.47(sin 66) = 4.084.08mm//ss

VVxx = 4.47(cos 66) = = 4.47(cos 66) = 1.82 1.82 mm//ss

= .83 sec= .83 sec

BB. max height =½t. max height =½t

y = Vy = Vyyt + ½ gtt + ½ gt22

y = (4.08)(.417) + ½ (-9.8)y = (4.08)(.417) + ½ (-9.8)(.417)(.417)22

y = 1.7 – 0.852y = 1.7 – 0.852y = 0.85my = 0.85m

CC. . Flight time Flight time = .833sec= .833sec

X = VX = Vxxt t

= 1.82m/s = 1.82m/s (.833sec)(.833sec)

X = 1.52mX = 1.52m

Practice problem 5

A softball is thrown with an initial velocity of 27m/s at an angle of 30o from the horizontal.

A. Find the total time the ball is in the air.

B. find the horizontal displacement of the ball.

C. Find the maximum height for the ball.

GivenGiven

VVxx = = 23.3823.38

VVyy = 13.5 = 13.5

g = -g = -9.8m/s9.8m/s

½ t= 1.38½ t= 1.38

Vyy3030°°

VVxx

27 m/s27 m/s

A. A. t = -2Vt = -2Vyy/ / gg

= -2(13.5)/- = -2(13.5)/- 9.89.8

= 2.76s= 2.76s

B.B. X = V X = Vxxtt

= = 23.38(2.76)23.38(2.76)

=65.5m=65.5mC.C. y = V y = Vyyt + ½ gtt + ½ gt22

= 13.5(1.38) + ½ (-9.8)= 13.5(1.38) + ½ (-9.8)(1.38)(1.38)22

18.63 + (-9.33)18.63 + (-9.33)

= 9.3m = 9.3m

Problem 6

All the given information of this problem is the same as #5. The only difference is the launch angle is now 60o.

A. find time B. Find range (X) C. Find max height (Y).

Vyy

VVxx

27 m/s27 m/s

6060°°

GivenGiven

VVxx = 13.5 = 13.5

VVyy = = 23.3823.38

g = -9.8g = -9.8

½ t = ½ t = 2.39s2.39s

A. A. t = -2Vt = -2Vyy/ g/ g

-2(23.38)/-g-2(23.38)/-g

= 4.78s= 4.78s

B.B. X = V X = Vxxt t

13.5(4.78)13.5(4.78)

=64.5m=64.5m

C.C. Y = V Y = Vyyt+ ½ gtt+ ½ gt22

= 23.38 (2.39) + ½ (-= 23.38 (2.39) + ½ (-9.8) 2.399.8) 2.3922

=27.9m=27.9m

The Range FormulaThe Range Formula

A fast way to find range when A fast way to find range when initial velocity and angle are initial velocity and angle are

given (angle should be with the given (angle should be with the horizontal)horizontal)

RR=VVii22 sin(2 0 ) sin(2 0 )

gg

Vi2 g Rg R

sin (2 0 )sin (2 0 )=Can Can also also

Find VFind Vii

ExampleExample Find the Find the rangerange for an object shot at for an object shot at

5050°° with an with an initial velocity of 5m/sinitial velocity of 5m/s

GivenGiven

VVii = 5m/s = 5m/s

0 = 50°0 = 50°

R = ?R = ?

g = 9.8g = 9.8

FormulaFormula

R = VR = Vii22 Sin 2 Sin 2

0 0

gg

SolutionSolution

= (5)= (5)22 sin sin (2x50(2x50°°) / 9.8) / 9.8

+ 24.62/ 9.8 = + 24.62/ 9.8 =

2.51m2.51m

p.104 3E #’s 5,2,1

p.114-115 30, 32, 34, 36, 39

Complimentary Angles = same ranges

450 gives the longest range for a projectile shot without air resistance. While the ranges may be the same, the time in the air won’t.

What are satellites?

Natural Artificial

How do satellites stay in orbit even though they have no

propulsion?

Isaac Newton saw the answer to this by thinking about the Moon. He also knew over 400 years ago that it would be possible to orbit the Earth if mankind could go fast enough. He also predicted mankind would travel to the Moon.

For low Earth orbit(200 to 500miles high)

Tangential Velocity of 8000m/s (17,500mph)

Vt will change with altitude.

The object (satellite) will stay in orbit. Every second it falls down 4.9m and

moves downrange (tangentially) 8000m. The resultant path mirrors the curvature of the Earth.

Vectors and Projectiles Review

Mr. Crabtree is standing on a perfectly good bridge in Zimbabwe when he decides to jump off. He is tied by his ankles to a bungee cord for safety. For 5.2 seconds, your teacher is in free fall. He jumped horizontally at 4 m/s as he left the bridge. The bridge is 150m above the Zambezi River. What was his vertical displacement at the end of the free fall? [neglect wind resistance]

Find horizontal displacement from the base of the building.

Find the time the ball is in the air

10m

20m