Physics 101 Lecture 4 Motion in 2D&3D (Projectile Motion) Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com
Physics 101Lecture 4Motion in 2D&3D
(Projectile Motion)Assist. Prof. Dr. Ali ÖVGÜN
EMU Physics Department
www.aovgun.com
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
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❑ 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)( ++=
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❑ 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
−=
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❑ 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 +=+==
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?
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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
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❑ 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 +=+==
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
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 +=
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❑ 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
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
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 +=
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
−=
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 +=
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=
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
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
++=
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)
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
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?
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Problem:3
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Problem:4
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Problem:5
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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.
February 5-8, 2013
Uniform circular motion
Constant speed, or,constant magnitude of velocity
Motion along a circle:Changing direction of velocity
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
=
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
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
⊥
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 .
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