monday bring rulers today and protractors!! wait till chapter 6...

Today

Monday

Monday/Weds

Monday

Wait till Chapter 6

Bring RULERSAnd Protractors!!

1-D Motion in a nutshell0

2

2

Averages: , , 2

Instantaneous: , ,

fv vx vv v at t

dx dv d xv a adt dt dt

+∆ ∆= = =∆ ∆

= = =

fv fv

0

20 0

2 20

Kinematics Eqs:

1 2

2

f

f

f

v v at

x x v t at

v v a x

= +

= + +

= + ∆

Constant acceleration.

0

( )t

f fv v a t dt= + ∫

00

( )t

fx x v t dt= + ∫

Varying acceleration.

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

Study for Monday!

212f i ir r v t at= + +

f iv v at= +

Projectile Motion: Vector Picture

212

= −f ir v gt t

Motion withno acceleration

ˆ ˆ( )= + −v xi yiv i v gt j

decelerationStudy for Monday!

Projectile MotionIgnoring Air Resistance

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 ∆y determines time!

Projectile MotionPerpendicular Directions are

INDEPENDENT

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!

Perpendicular Directions are INDEPENDENT

No Change

change

0xa =

ya g=

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

Projectile Motion

The y-component is in FREE FALL! changes (a = g.)

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)

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 20∆ = = =ximx v t s ms

210 /g m s=

212yiy v t gt∆ = + 2 1yt s

g∆

= =0

Question

d

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


30 1 30∆ = = =ximx v t s ms

210 /g m s=

212yiy v t gt∆ = + 2 1yt s

g∆

= =

Only ∆y determines time!

QuestionThe ball is thrown horizontally at 100 m/s.

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


100 1 100∆ = = =ximx v t s ms

210 /g m s=

Curvature of EarthCurvature of the Earth: Every 8000 m,

the Earth curves by 5 meters!

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 8000∆ = = =xhorizontal x v t m s s m

( )22 21: 5 5 1 52

∆ = = = =vertical y gt t s m




Does the ball ever hit the Earth????

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.

Projectile Motion IS Orbital MotionThe Earth is in the way!

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 Problem Solving•IGNORE AIR RESISTANCE!!•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!!!!

Most Important:

X and Y

components are

INDEPENDENT

of each other!

Perpendicular Directions are INDEPENDENT

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

Only the vertical component changes!

First the SIMPLE Case: Horizontal Launch



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?

?x∆ =

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 t∆ =

Care Package

0

2

:0, 115 /1050 , ?9.8 / , 0

y x

y x

Knownsv v m s

y m xa m s a

= =

∆ = − ∆ =

= − =

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

20

1 22 y

y

yy v t a t ta∆

∆ = + → =

∆ = xx v t

115 / (14.6 )= m s s

1680x m∆ =

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

= =

∆ = − ∆ =

= − =

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

− − − = =

= −

(184 / , 51.3 )v m s= −

Projectiles Launched at an Angle:The simple case: ∆y=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.

Maximum range is achieved at a launch angle of 45°!

Symmetry in the Projectile Rangeis symmetric about 45°

2 sin 2i ivRg

θ=

sin 2θRange Equation:

Projectile Motion Launched at an Angle



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.

A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53degrees above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

How Tall is the Building?

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

∆ = +y yy v t a t

0∆ = xx v t cos50 10 cos15=

md ts

22(15.03 )sin50 10 sin15 4.9m mt t t

s s− = −

2.88t s=

cos1510 15.03cos50

md t ts

= =

22sin 50 10 sin15 4.9− = −

m md t ts s

211.51 2.59 4.9 m ts

− = − 27.8 , 43.2x m d m∆ = =

QuizProblem : A dive bomber has a velocity of 280 m/s at an angle below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25 km. Find the angle .

HINT:

You don’t know time so set up two equations for ∆y and ∆x, eliminate t and use the displacement values given to solve for θ.

monday bring rulers today and protractors!! wait till chapter 6...

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