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Physics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com
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Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

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Page 1: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

Physics 101Lecture 4Motion in 2D&3D

(Projectile Motion)Assist. Prof. Dr. Ali ÖVGÜN

EMU Physics Department

www.aovgun.com

Page 2: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Vector and its components

❑ The components are the

legs of the right triangle

whose hypotenuse is Ayx AAA

+=

2 2 1tany

x y

x

AA A A and

A −

= + =

)sin(

)cos(

=

=

AA

AA

y

x

Or,

( ) ( )

( )

==

+=

x

y

x

y

yx

A

A

A

A

AAA

1

22

tanor tan

Page 3: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Kinematic variables in one dimension

◼ Position: x(t) m

◼ Velocity: v(t) m/s

◼ Acceleration: a(t) m/s2

❑ Kinematic variables in three dimensions

◼ Position: m

◼ Velocity: m/s

◼ Acceleration: m/s2

❑ All are vectors: have direction and

magnitudes

Motion in two dimensions

kvjvivtv zyxˆˆˆ)( ++=

y

x

z

ij

k

x

kzjyixtr ˆˆˆ)( ++=

kajaiata zyxˆˆˆ)( ++=

Page 4: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ In one dimension

❑ In two dimensions

◼ Position: the position of an object is

described by its position vector -

-always points to particle from origin.

◼ Displacement:

x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 mΔx = +1.0 m + 3.0 m = +4.0 m

Position and Displacement

)(tr

12 rrr

−=

jyix

jyyixx

jyixjyixr

ˆˆ

ˆ)(ˆ)(

)ˆˆ()ˆˆ(

1212

1122

+=

−+−=

+−+=

)()( 1122 txtxx −=

12 rrr

−=

Page 5: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Average velocity

❑ Instantaneous velocity

❑ v is tangent to the path in x-y graph;

Average & Instantaneous Velocity

dt

rd

t

rvv

tavg

=

=

→→ 00tlimlim

jvivjt

yi

t

xv yavgxavgavg

ˆˆˆˆ,, +=

+

=

t

rvavg

jvivjdt

dyi

dt

dx

dt

rdv yx

ˆˆˆˆ +=+==

Page 6: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Motion of a Turtle

A turtle starts at the origin and moves with the speed of v0=10 cm/s in

the direction of 25° to the horizontal.

(a) Find the coordinates of a turtle 10 seconds later.

(b) How far did the turtle walk in 10 seconds?

Page 7: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Motion of a Turtle

Notice, you can solve the

equations independently for the

horizontal (x) and vertical (y)

components of motion and then

combine them!

yx vvv

+=0

0 0 cos 25 9.06 cm/sxv v= =

❑ X components:

❑ Y components:

❑ Distance from the origin:

0 90.6 cmxx v t = =

0 0 sin 25 4.23 cm/syv v= = 0 42.3 cmyy v t = =

cm 0.10022 =+= yxd

Page 8: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Average acceleration

❑ Instantaneous acceleration

❑ The magnitude of the velocity (the speed) can change

❑ The direction of the velocity can change, even though the

magnitude is constant

❑ Both the magnitude and the direction can change

Average & Instantaneous Acceleration

dt

vd

t

vaa

tavg

=

=

→→ 00tlimlim

jaiajt

vi

t

va yavgxavg

yxavg

ˆˆˆˆ,, +=

+

=

t

vaavg

jaiajdt

dvi

dt

dv

dt

vda yx

yx ˆˆˆˆ +=+==

Page 9: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Position

❑ Average velocity

❑ Instantaneous velocity

❑ Acceleration

❑ are not necessarily same direction.

Summary in two dimensionjyixtr ˆˆ)( +=

jaiajdt

dvi

dt

dv

dt

vd

t

vta yx

yx

t

ˆˆˆˆlim)(0

+=+==

=

jvivjt

yi

t

x

t

rv yavgxavgavg

ˆˆˆˆ,, +=

+

=

=

jvivjdt

dyi

dt

dx

dt

rd

t

rtv yx

t

ˆˆˆˆlim)(0

+=+==

=

dt

dxvx

dt

dyvy

2

2

dt

xd

dt

dva x

x =2

2

dt

yd

dt

dva

y

y =

)( and ),( , tatv(t)r

Page 10: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Motion in two dimensions

tavv

+= 0

❑ Motions in each dimension are independent components

❑ Constant acceleration equations

❑ Constant acceleration equations hold in each dimension

◼ t = 0 beginning of the process;

◼ where ax and ay are constant;

◼ Initial velocity initial displacement ;

2

21

0 tatvrr

+=−

tavv yyy += 0

2

21

00 tatvyy yy +=−

)(2 0

2

0

2yyavv yyy −+=

tavv xxx += 0

2

21

00 tatvxx xx +=−

)(2 0

2

0

2xxavv xxx −+=

jaiaa yxˆˆ+=

jvivv yxˆˆ

000 +=

jyixr ˆˆ000 +=

Page 11: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 12: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 13: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 14: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ 2-D problem and define a coordinate

system: x- horizontal, y- vertical (up +)

❑ Try to pick x0 = 0, y0 = 0 at t = 0

❑ Horizontal motion + Vertical motion

❑ Horizontal: ax = 0 , constant velocity

motion

❑ Vertical: ay = -g = -9.8 m/s2, v0y = 0

❑ Equations:

Projectile Motion

2

21 gttvyy iyif −+=tavv yyy += 0

2

21

00 tatvyy yy +=−

)(2 0

2

0

2yyavv yyy −+=

tavv xxx += 0

2

21

00 tatvxx xx +=−

)(2 0

2

0

2xxavv xxx −+=

Horizontal Vertical

Page 15: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ X and Y motions happen independently, so

we can treat them separately

❑ Try to pick x0 = 0, y0 = 0 at t = 0

❑ Horizontal motion + Vertical motion

❑ Horizontal: ax = 0 , constant velocity

motion

❑ Vertical: ay = -g = -9.8 m/s2

❑ x and y are connected by time t

❑ y(x) is a parabola

Projectile Motion

gtvv yy −= 0

2

21

00 gttvyy y −+=

xx vv 0=

tvxx x00 +=

Horizontal Vertical

Page 16: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ 2-D problem and define a coordinate system.

❑ Horizontal: ax = 0 and vertical: ay = -g.

❑ Try to pick x0 = 0, y0 = 0 at t = 0.

❑ Velocity initial conditions:

◼ v0 can have x, y components.

◼ v0x is constant usually.

◼ v0y changes continuously.

❑ Equations:

Projectile Motion

000 cosvv x =

Horizontal Vertical

000 sinvv x =

gtvv yy −= 0

2

21

00 gttvyy y −+=

xx vv 0=

tvxx x00 +=

Page 17: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Initial conditions (t = 0): x0 = 0, y0 = 0

v0x = v0 cosθ0 and v0y = v0 sinθ0

❑ Horizontal motion:

❑ Vertical motion:

❑ Parabola;

◼ θ0 = 0 and θ0 = 90 ?

Trajectory of Projectile Motion

2

21

00 gttvy y −+=

x

xv

xttvx

0

0 0 =+=

2

00

02

=

xx

yv

xg

v

xvy

2

0

22

0

0cos2

tan xv

gxy

−=

Page 18: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Initial conditions (t = 0): x0 = 0, y0 = 0

v0x = v0 cosθ0 and v0x = v0 sinθ0, then

What is R and h ?

Horizontal Vertical

2

21

000 gttv y −+=tvx x00+=

g

v

g

vvtvxxR x

0

2

0000000

2sinsincos2 ===−=

g

v

g

vt

y 000 sin22 ==

2

0

2

21

00222

−=−=−=

tgtvgttvyyh yhhy

g

vh

2

sin 0

22

0 =

y

y

yyy vg

vgvgtvv 0

0

00

2−=−=−=

h

gtvv yy −= 0

2

21

00 gttvyy y −+=

xx vv 0=

tvxx x00 +=

Page 19: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Projectile Motion

at Various Initial Angles

❑ Complementary

values of the initial

angle result in the

same range

◼ The heights will be

different

❑ The maximum range

occurs at a projection

angle of 45o

g

vR

2sin2

0=

Page 20: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Position

❑ Average velocity

❑ Instantaneous velocity

❑ Acceleration

❑ are not necessarily in the same direction.

Summaryjyixtr ˆˆ)( +=

jaiajdt

dvi

dt

dv

dt

vd

t

vta yx

yx

t

ˆˆˆˆlim)(0

+=+==

=

jvivjt

yi

t

x

t

rv yavgxavgavg

ˆˆˆˆ,, +=

+

=

=

jvivjdt

dyi

dt

dx

dt

rd

t

rtv yx

t

ˆˆˆˆlim)(0

+=+==

=

dt

dxvx

dt

dyvy

2

2

dt

xd

dt

dva x

x =2

2

dt

yd

dt

dva

y

y =

)( and ),( , tatv(t)r

Page 21: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ If a particle moves with constant acceleration a, motion

equations are

❑ Projectile motion is one type of 2-D motion under constant

acceleration, where ax = 0, ay = -g.

Summary

jtatvyitatvxjyixr yiyiixixiifffˆ)(ˆ)(ˆˆ 2

212

21 +++++=+=

jtavitavjvivtv yiyxixfyfxfˆ)(ˆ)(ˆˆ)( +++=+=

tavv i

+=

2

21 tatvrr iif

++=

Page 22: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

Example: 1

February 5-8, 2013

Example: 2

A movie stunt driver on a motorcycle speeds horizontally off a

50m high cliff. If the motorcycle will land 90m from the base

of the cliff, (ignore any kind of friction or resistance) (a) Find

the time of flight, (b) Find its initial speed in x-direction ,

(c) Find its acceleration vector just before

hitting the ground.

During volcanic eruptions, chunks of solid rock can be blasted out of the

volcano; these projectiles are called volcanic bombs. The figure below shows a

cross section of Mt. Fuji, in Japan. From the vent A to the foot of the volcano at

B, the vertical distance is h = 3.30km and horizontal distance is d = 940m.

Neglecting air resistance,

(a) calculate the time of flight, and (4 P)

(b) calculate the initial speed of the projectile. (2P)

Page 23: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

A cliff diver is about to jump down a

cliff of height 35.0m, at the bottom of

the cliff there is a 5m wide rock bank

next to the sea. Calculate the

minimum horizontal initial velocity the

cliff jumper has to push off. (No initial

velocity component in y direction)

Example: 3

Page 24: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

Problem:1

February 5-8, 2013

A projectile is fired at an initial velocity of 35.0

m/s at an angle of 30.0 degrees above the

horizontal from the roof of a building 30.0 m

high, as shown. Find

a) The maximum height of the projectile

b) The time to rise to the top of the trajectory

c) The total time of the projectile in the air

d)The velocity of the projectile at the ground

e)The range of the projectile

Problem:2

aA plane drops a package

of supplies to a party of

explorers. If the plane is

traveling horizontally at 40

m/s and is 100 m above the

ground. Where does the

package strike the ground?

Page 25: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Problem:3

Page 26: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Problem:4

Page 27: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Problem:5

Page 28: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Problem:6

At what initial speed must the basketball player in

Figure throw the ball, at angle u0 = 37° above the

horizontal, to make the foul shot?

The horizontal distances are = 30 cm and

, = 440cm and the heights are = 220 cm

and . = 300 cm.

Page 29: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Uniform circular motion

Constant speed, or,constant magnitude of velocity

Motion along a circle:Changing direction of velocity

Page 30: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Circular Motion: Observations

❑ Object moving along a curved path with constant speed

◼ Magnitude of velocity: same

◼ Direction of velocity: changing

◼ Velocity: changing

◼ Acceleration is NOT zero!

◼ Net force acting on the object is NOT zero

◼ “Centripetal force” amFnet

=

Page 31: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

❑ Centripetal acceleration

❑ Direction: Centripetal

Uniform Circular Motion

r

v

t

va

r

v

r

v

t

r

t

v

r

rvv

r

r

v

v

r

2

2

so,

=

=

=

=

=

=

Ox

y

ri

R

A Bvi

rf

vf

Δr

vi

vf

Δv = vf - vi

Page 32: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Uniform Circular Motion

❑ Velocity:◼ Magnitude: constant v

◼ The direction of the velocity is tangent to the circle

❑ Acceleration:◼ Magnitude:

◼ directed toward the center of the circle of motion

❑ Period: ◼ time interval required for one

complete revolution of the particle

r

vac

2

=

r

vac

2

=

v

rT

2=

vac

Page 33: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Problem:8

The bobsled track contains turns

with radii of 33 m and 24 m.

Find the centripetal acceleration

at each turn for a speed of

34 m/s. Express answers as

multiples of .

Page 34: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 35: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 36: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013

Page 37: Physics 101 Lecture 4 Motion in 2D&3DPhysics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department . February 5-8, 2013 Vector and

February 5-8, 2013