Projectile Motion

2-D Vector Equations have the same form as 1-D Kinematics

ˆ ˆ( ) ( )xi x yi yv a t i v a t j= + + +

ˆ ˆf x yv v i v j= +v

ˆ ˆ ˆ ˆ( ) ( )xi yi x yv i v j a ti a tj= + + +

f iv v at= +v v v

212f i ir r v t at= + +v v v v

Similarly,

Two Dimensional Vector Motion

ˆ ˆr xi yj= +vˆ ˆdr dx dyv i j

dt dt dt= = +

vv

ˆ ˆx yv v i v j= +v

2-D Vector Equations have the same form as 1-D Kinematics

f iv v at= +v v v 212f i ir r v t at= + +v v v v

Projectile Motion: Vector Picture

212f ir v t gt= −v v v

Motion withno acceleration

ˆ ˆ( )xi yiv v i v gt j= + −v

Vector Problem

A particle initially located at the origin has an acceleration of       

and an initial velocity of  

Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t = 2.00 s.

a = 3.00ˆ j m / s 2

v i = 5.00ˆ i m / s

t= +v v af i21

2t t= + +f i ir r v a

Projectile MotionIgnore Air Resistance!

Most Important:X and Y components are INDEPENDENT of each other!

Zero at the Top!Y component of velocity is zero at the top of the path in both cases!

You know they have the y component of velocity is the same in both cases because they reached the same height!

Projectile Motion

First the SIMPLE Case: Horizontal Launch

The x-component doesn’t change (no acceleration in x-direction.)The y-component changes (a = -g.)

(Ignore Air Resistance)

Horizontal and vertical components are independent of each other!

Gravity acts in the vertical direction but not in the horizontal direction!!

Speed in vertical direction speeds up!Speed in horizontal direction stays the same!

Projectile Motion

No Change

change

0xa =

ya g=

Actual path is a vector sum of horizontal and vertical motions.

Projectile Motion

Plane and PackageAn airplane traveling at a constant speed and height drops a care package. Ignoring air resistance, at the moment the package hits the ground, where is it relative to the plane?

a) Behind the plane.b) Under the plane.c) In front of the plane.

Dropping From Moving Frame

Any object dropped from a plane has the same initial velocity as the plane!

Care PackageAn airplane moves horizontally with constant velocity of 115 m/s

at an altitude of 1050m and drops a care package as shown. How far from the release point does the package land?

?xΔ =

1050m

Care PackageStrategy: Find the time the package drops to get the horizontal

distance. The time to drop is just the free fall time!!! The horizontal displacement takes the same time as it takes the

package to drop.

xx v tΔ =

Care Package

0

2

:0, 115 /

1050 , ?9.8 / , 0

y x

y x

Knownsv v m s

y m xa m s a

= =

Δ = − Δ =

= − =

Strategy: Find time from y info to solve for .xx v tΔ =

20

1 22 y

y

yy v t a t taΔ

Δ = + → =

Δ = xx v t

115 / (14.6 )= m s s

1680x mΔ =

14.6t s=

With what velocity does it hit the ground?

2

2( 1050 )( 9.8 / )−

=−

mtm s

Care Package

0

2

:0, 115 /

1050 , 16809.8 / , 0

y x

y x

Knownsv v m s

y m x ma m s a

= =

Δ = − Δ =

= − =

Strategy: Find final velocity in y direction and use it in:

2 2 1, tan yx y

x

vv v v

vθ − ⎛ ⎞

= + = ⎜ ⎟⎝ ⎠

0yf y yv v a t= +14.6t s= 20 ( 9.8 / )(14.6 )m s s= + −

143 /m s= −

2 2

2 2(115 / ) ( 143 / )184 /

x yv v v

m s m sm s

= +

= + −

=

1 1 143 /tan tan115 /

51.3

y

x

v m sv m s

θ

θ

− −⎛ ⎞ −⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= − o(184 / , 51.3 )v m s= − ov

Projectile MotionSame cannons, Same height. One dropped, One shot.

Which hits the ground first? SAME!Both falling the same height!

Horizontal speed doesn’t affect vertical speed or the time to hit the ground!

Only Δy determines time!

Question

d

The ball is thrown horizontally at 20 m/s.About how long does it take to hit the ground?

How far does it travel in the horizontal direction?

20 1 20Δ = = =ximx v t s ms

210 /g m s=

212yiy v t gtΔ = + 2 1yt s

gΔ

= =0

Question

d

The ball is thrown horizontally at 30 m/s.About how long does it take to hit the ground?

How far does it travel in the horizontal direction?

30 1 30Δ = = =ximx v t s ms

210 /g m s=

212yiy v t gtΔ = + 2 1yt s

gΔ

= =

Only Δy determines time!

QuestionThe ball is thrown horizontally at 100 m/s.

How long does it take to hit the ground? 1 Second!!

How far does it travel in the horizontal direction?

100 1 100Δ = = =ximx v t s ms

210 /g m s=

Curvature of EarthIf you threw the ball at 8000 m/s off the surface of the Earth

(and there were no buildings or mountains in the way)how far would it travel in the vertical and horizontal

directions in 1 second?

Curvature of EarthIf you threw the ball at 8000 m/s off the surface of the Earth

(and there were no buildings or mountains in the way)how far would it travel in the vertical and horizontal

directions in 1 second?

( )( ): 8000 / 1 8000Δ = = =xhorizontal x v t m s s m

( )22 21: 5 5 1 52

Δ = = = =vertical y gt t s m

Curvature of EarthIf you threw the ball at 8000 m/s off the surface of the Earth

(and there were no buildings or mountains in the way)how far would it travel in the vertical and horizontal

directions in 1 second?

( )( ): 8000 / 1 8000Δ = = =xhorizontal x v t m s s m

( )22 21: 5 5 1 52

Δ = = = =vertical y gt t s m

Curvature of EarthIf you threw the ball at 8000 m/s off the surface of the Earth

(and there were no buildings or mountains in the way)how far would it travel in the vertical and horizontal

directions in 1 second?

Does the ball ever hit the Earth????

Curvature of EarthCurvature of the Earth: Every 8000 m,

the Earth curves by 5 meters!

Orbital VelocityIf you can throw a ball at 8000m/s, the Earth curves away

from it so that the ball continually falls in free fall around the Earth – it is in orbit around the Earth!

Above the atmosphere

Ignoring air resistance.

Orbital Motion| & Escape Velocity8km/s: Circular orbit

Between 8 & 11.2 km/s: Elliptical orbit11.2 km/s: Escape Earth

42.5 km/s: Escape Solar System!

Projectile Motion IS Orbital MotionThe Earth is in the way!

Projectile Motion Problem Solving•x and y directions are INDEPENDENT•INDEPENDENT kinematics equations for x and y direction•gravity affects only y direction•solve for x with a = 0! Use a = g for y direction only!•time for events to occur is the same for x and y directions•time is the link between x and y!!!!

Same rock, same speed, same angle.Which rock hits the water first?

a) Rock 1 b) Rock 2 c) same

Which rock hits the water with the greatest speed?a) Rock 1 b) Rock 2 c) same

Which rock hits the water first?a) Rock 1 b) Rock 2 c) same

Which rock hits the water with the greatest speed?a) Rock 1 b) Rock 2 c) same

SpatialSymmetryIn G Field!

Same rock, same speed, same angle.

Projectile Motion Launched at an Angle

The x-component doesn’t change (no acceleration in x-direction.)The y-component changes (a = -g.)

(Ignore Air Resistance)

Projectiles Launched at an Angle:The simple case: Δy=0

Projectile MotionA place kicker kicks a football at an angle of 40 degrees above the horizontal with an initial speed of 22 m/s. Ignore air resistance and find the total time of flight, the maximum height and the range the

ball attains.

Projectiles Launched at an Angle:The simple case: Δy=0

2 2sin2

i ivhgθ

=2 sin 2i ivR

gθ

=

Range and Maximum Height of a Projectile: WARNING

Only Good For This Motion

2 2sin2

i ivhgθ

=

2 sin 2i ivRg

θ=

Not Good for Non-Symmetric Projectile Motion

2 2sin2

i ivhgθ

=

2 sin 2i ivRg

θ=

Not Good for Non-Symmetric Projectile Motion

2 2sin2

i ivhgθ

=

2 sin 2i ivRg

θ=

Symmetry in the Projectile Rangeis symmetric about 45°

2 sin 2i ivRg

θ=

sin 2θRange Equation:

Study Sample Problem!

f xix v t=GET t from y info!!!

212yi yy v t a tΔ = +

2solve( 45 20sin 30 4.9 )t t− = −o

4.22t s=20 / cos30 (4.22 )fx m s s= o

73.1fx m=

What is the Range of motion?

Harder Horizontal Launch Problems: Hitting an incline!

Distance traveled is given by the trajectory but the net displacement is the diagonal!

A skier leaves the ramp of a ski jump with a velocity of 10.0 m/s, 15.0° above the horizontal, as shown.  The slope is inclined at 50.0°, and air resistance is negligible. Find the distance from the ramp to where the jumper lands and the time of flight.

dy

x

20

12

Δ = +y yy v t a t

0Δ = xx v t cos50 10 cos15=o omd ts

22sin 50 10 sin15 4.9− = −o om md t t

s s

Solving simultaneously:

43.2 , 2.88= =d m t s

Car ProblemA car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37.0° below the horizontal.  The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.00 m/s2, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean.  Find 

(a) the speed of the car when it reaches the edge of the cliff and the time at which it arrives there, 

(b) the velocity of the car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the position of the car when it lands in the ocean, relative

to the base of the cliff.

Galilean RelativityNewton’s Principia in 1687.

The The GalileanTransformationsGalileanTransformationsConsider two reference frames S and S'. The coordinate axes in S are x, y, z and those in S' are x', y', z'. Reference frame S'moves with velocity v relative to S along the x-axis. Equivalently, S moves with velocity −v relative to S'. The Galilean transformations of position are:

The Galilean transformations of velocity are:

Shooting UP From Moving Frame

Horizontal and vertical components are independent of each other!Horizontal component remains unchanged without air resistance.

Only the vertical component changes!

Relative Velocity• Two observers moving relative to each other generally do

not agree on the outcome of an experiment• For example, observers A and B below see different paths

for the ball and measure different velocities:

= +v v vbB bA ABv v vVelocity of ball

relative to observer B

Velocity of ball relative to observer A

Velocity of A relative to observer B

Co-Linear MotionJust add or subtract the magnitudes of vectors!

PG PT TGv v v= +

Notice how the inner subscripts cancel!

2D Relative VelocityThe boat can travel 2.50 m/s relative to the river. The river current flows at 1.00 m/s relative to the Earth. What is the total velocity of the boat relative to the Earth (shore = Earth)?

= +v v vbE br rEv v v

2.50 /m s

1 /m s2 2(2.5 / ) (1 / ) 2.69 /= + =bEv m s m s m s

1 1m /stan 21.82.5m /s

− ⎛ ⎞= =⎜ ⎟⎝ ⎠

oθ

Relative Velocity AgainThe boat can travel 10 m/s relative to the river. The river current flows at 5.00 m/s relative to the shore. If the boat wants to travel straight across, what must be his heading? What is its total speed?

= +v v vbE br rEv v v

1 5 /sin 3010 /

m sm s

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

o

From the triangle:

10 /m s

5 /m s

2 2(10 / ) (5 / ) 8.66 /= − =bEv m s m s m s

2 2 2= +br bE rEv v v

Train RainA person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 5.00 m/s relative to the ground. When the train moves at a constant velocity, the rain drops make an angle of 25 degrees when the move past the window, as the drawing shows. How fast is the train moving?

RT RG GTv v v= +We know:

We want vTG:

RG TGv v= − 25oRGv

TGv−

RTv

tan 25TG RGv v= o

5 / tan 25m s= o 2.33 /m s=

You Try RainA car travels in a due northerly direction

at a speed of 55 km/h. The traces of rain on the side windows of the car make an angle of 60 degrees with respect to the horizontal. If the rain is falling vertically with respect to the earth, what is the speed of the rain with respect to the earth?a. 48 km/hb. 95 km/hc. 58 km/hd. 32 km/he. 80 km/h

While driving a car in the rain falling straight down relative to the ground, the rear window can remain dry! Why?

We know the velocity of the raindrop and the car relative to the ground. To determine whether the raindrop hits the window we need to consider the velocity of the raindrop relative to the car:

RC RG GCv v v= +

RG CGv v= −

The velocity of the ground relative to the car is just negative

the velocity of the carrelative to the ground!

If the direction of the rain relative to the car, θR, is greater than the angle of the rear window, θW, the rain will not hit the rear window! The faster the car, the greater the angle of the rain and the rear window can remain dry!

1tan ( )CGR

RG

vv

θ −=

From the vector diagram of the relative velocities:

ØW

ØR

Previous problems all involved right triangles….how do you solve if you don’t have right triangle relationship

between relative velocities?

Using ijk vector component additionA ferry boat is traveling in a direction 35.1 degrees north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71m/s due east relative to the boat. What is the velocity of the passenger with respect to the water? Determine the angle relative to due east.

2 2(6.9) (2.94) / 7.50 /PWv m s m s= + =

PWv

PBv

BWv

ˆ ˆ ˆ[5.12cos(35.1) 5.12sin(35.1) ] / (2.71 ) /i j m s i m s= + +

PW PB BWv v v= +

ˆ ˆ(6.90 2.94 ) /i j m s= +

1 2.94tan ( ) 23.16.9PWθ −= = o

PWθ

Using Law of Sines and CosinesA ferry boat is traveling in a direction 35.1 degrees north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71m/s due east relative to the boat. What is the velocity of the passenger with respect to the water? Determine the angle relative to due east.

PWv

PBv

BWvPW PB BWv v v= +

35.1o

35.1o

144.9o

2 2 2

Low of cosines: 2 cos 20= + −

o

c a b ab

2 2 2 2 cos144.9= + − oPW BW PB BW PBv v v v v

7.50 /PWv m s=2 2(5.12 2.71 2(5.12)(2.71)cos144.9 ) /= + − oPWv m s

sin sin sinLaw of sines: a b cα β γ= =

sin sin144.92.71 7.5α=

o

α

11.99α = o 35.1 11.99 23.1PWθ = − =o o o 23.1PWθ = o

PWθ

Using ijk vector component additionA ferry boat is traveling in a direction 35.1 degrees northof east with a speed of 5.12 m/s relative to the water. Apassenger is walking with a velocity of 2.71m/s due eastrelative to the boat. What is the velocity of the passengerwith respect to the water? Determine the angle relative todue east.

2 2(6.9) (2.94) / 7.50 /PWv m s m s= + =

PWv

PBv

BWv

ˆ ˆ ˆ[5.12cos(35.1) 5.12sin(35.1) ] / (2.71 ) /i j m s i m s= + +

ˆ ˆ(6.90 2.94 ) /i j m s= +

PW PB BWv v v= +

1 2.94tan ( ) 23.16.9PWθ −= = �

Using Law of Sines and CosinesA ferry boat is traveling in a direction 35.1 degrees northof east with a speed of 5.12 m/s relative to the water. Apassenger is walking with a velocity of 2.71m/s due eastrelative to the boat. What is the velocity of the passengerwith respect to the water? Determine the angle relative todue east.

PWv

PBv

BWvPW PB BWv v v= +

35.1�

35.1�

144.9�

2 2 2Low of cosines: 2 cos 20= + − �c a b ab

2 2 2 2 cos144.9= + − �PW BW PB BW PBv v v v v

7.50 /PWv m s=2 2(5.12 2.71 2(5.12)(2.71)cos144.9 ) /= + − �PWv m s

sin sin sinLaw of sines: a b cα β γ= =

sin sin144.92.71 7.5α=

�

α

11.99α = � 35.1 11.99 23.1PWθ = − =� � � 23.1PWθ = �

Which Way is Best????

Depending on the givens, COMPONENTS IS EASIER!!!AND it works for more than three vectors!!!!

You Try: Relative VelocityA plane is moving at 45m/s due north relative to the air, while its

velocity relative to the ground is 38.0m/s, 20 degrees west of north.What is the velocity of the wind relative to due west?

/ / /= +v v vplane ground plane wind wind groundv v v

2 2 2 2 cos 20c a b ab= + − o

One Method: Use Law of Cosines:

2 238 45 2(38)(45)cos 20c = + − o

16 /wv m s=

1 38sin(20) /cos 35.716 /

m sm s

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

o

#34 Vector DriversWarning: Typo!!! Not 300!

Heather in her Corvette accelerates at the rate of

while Jill in her Jaguar accelerates at

They both start from rest at the origin of an xycoordinate system.  After 5.00 s, find 

• the velocity of each driver and Heatherʹs speed with respect to Jill.

• The displacement vector for each driver and how far apart are they, and draw it all.

• What is Heatherʹs acceleration relative to Jill?

2ˆ ˆ(3.00 2.00 ) /i j m s−

2ˆ ˆ(1.00 3.00 ) / .i j m s+

Relative RainA train travels due south at 30m/s (relative to the ground) in a rain that is blown toward the south by the wind blowing due south. The path of each raindrop makes an angle of 70 degrees with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops.

A coast guard ship is traveling at a constant velocity of 4.20 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2310 m with respect to the ship, in a direction 32 degrees south of east. Six minutes later, he notes that the object’s position relative to the ship has changed to 1120 m, 57 degrees south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

Hard Problem