Lecture 5
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Homework 4: The Plane
A plane flies 483 km west from city A to city B in 45 min and
then 966 km south from city B to city C in 1.5 h.
From the total trip of the plane, determine:
(a) the magnitude of its displacement;
(b) the direction of its displacement;
(c) the magnitude of its average velocity;
(d) the direction of its average velocity;
(e) its average speed.
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University Physics: Mechanics
5/2
Solution of Homework 4: The Plane
→
Δr1
B
A
 r1   4 8 3iˆ k m
 r   966 ˆj km
 t1  45 min  0.75 h
 t 2  1.5 h
2
966 km,
1.5 h
483 km,
45 min
Δr→2
 rtotal   r1   r2   483iˆ  966 ˆj km
 t total   t1   t 2  0 .7 5  1 .5  2 .2 5 h
(a) the magnitude of its displacement
 rto tal 
(  4 8 3)  (  9 6 6 )
2
2
 1080.021 km
2 4 3 .4 3 5 
B
A
(b) the direction of its displacement
C
  rtotal  tan
1
  966 


  483 
 6 3 .4 3 5 
• Quadrant I
C
 243.435  • Quadrant III
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5/3
Solution of Homework 4: The Plane
(c) the magnitude of its average velocity
v avg 
v avg 
 rtotal

 t total
 483iˆ  966 ˆj km
  214.667 ˆi  429.333 ˆj km h
2.25 h
(  2 1 4 .6 6 7 )  (  4 2 9 .3 3 3)
2
2
 480 km h
(d) the direction of its average velocity
 v avg  tan
1
  429.333   2 4 3 .4 3 5  • Quadrant III


  214.667 
(e) its average speed
s avg 
to ta l d ista n ce tra ve le d
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to ta l tim e

483 km  966 km
 644 km h
2.25 h
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5/4
Average and Instantaneous Acceleration
→
→
 When a particle’s velocity changes from
v1 to v2 in a time
→
interval Δt, its average acceleration aavg during Δt is:
a v e ra g e  ch a n g e in v e lo city
a cce le ra tio n
tim e in te rv a l
a avg 
v 2  v1
t

v
t
 If we shrink Δt to zero, then→→
aavg approaches the
instantaneous acceleration a ; that is:
a 
dv
dt
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dv x ˆ dv y ˆ dv z ˆ
ˆ
ˆ
ˆ

( v x i  v y j  v z k) 
i
j
k
dt
dt
dt
dt
d
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5/5
Average and Instantaneous Acceleration
 We can rewrite the last equation as
a  a x ˆi  a y ˆj  a z kˆ
→
where the scalar components of a are:
ax 
dv x
dt
, ay 
dv y
dt
, az 
dv z
dt
Acceleration of a particle does not have
to point along the path of the particle
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5/6
Average and Instantaneous Acceleration
→
A particle with velocity→v0 = –2i^ + 4j^ m/s at t = 0 undergoes a
constant acceleration a of magnitude a = 3 m/s2 at an angle
130° from the positive
direction of the x axis. What is the
→
particle’s velocity v at t = 5 s?
Solution:
v y  v0 y  a y t
v x  v0 x  a x t
v0 x   2 m s
v0 y  4 m s
a y  3  sin 130 
a x  3  cos130 
  1.928 m s
2
 2.298 m s
2
At t = 5 s,
v x   2  (  1.928)(5)
  1 1 .6 4 m s
v y  4  (2.298)(5)
 1 5 .4 9 m s
Thus, the particle’s velocity at t = 5 s is
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v   11.64iˆ  15.49 ˆj m s.
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5/7
Projectile Motion
 Projectile motion: a motion in a vertical plane, where
the
→
acceleration is always the free-fall acceleration g, which is
downward.
 Many sports involve the projectile motion of a ball.
 Besides sports, many acts also involve the projectile motion.
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5/8
Projectile Motion
 Projectile motion consists of horizontal motion and vertical
motion, which are independent to each other.
 The horizontal motion has no acceleration (it has a constant
velocity).
 The vertical motion is a free fall motion with constant
acceleration due to gravitational force.
ax  0
g  9.81 m s
2
ay  g
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University Physics: Mechanics
5/9
Projectile Motion
ax  0
v 0  v 0 x ˆi  v 0 y ˆj
ay  g
v 0 x  v 0 cos  0
v 0 y  v 0 sin  0
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5/10
Projectile Motion
Two Golf Balls
v0 x  0
v0 x  0
• The vertical motions are quasiidentical.
• The horizontal motions are
different.
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University Physics: Mechanics
5/11
Projectile Motion Analyzed
The Horizontal Motion
x  x0  v0 x t
x  x 0  ( v 0 cos  0 ) t
The Vertical Motion
y  y 0  v0 y t 
1
gt
2
y  y 0  ( v 0 sin  0 ) t 
1
2
2
gt
2
v y  v 0 sin  0  gt
v y  ( v 0 sin  0 )  2 g ( y  y 0 )
2
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2
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5/12
Projectile Motion Analyzed
The Horizontal Range
x  x0  R
R  ( v 0 cos  0 ) t
y  y0  0
0  ( v 0 sin  0 ) t 
vx = v0x
vy = –v0y
1
2
gt
2
Eliminating t,
2
R 2
v0
g
sin  0 co s  0
• This equation is valid if the landing
height is identical with the launch height.
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University Physics: Mechanics
5/13
Projectile Motion Analyzed
Further examining the equation,
2
v0
R 2
g
sin  0 co s  0
Using the identity
sin 2 0  2 sin  0 cos  0 ,
we obtain
2
R 
v0
g
sin 2 0
R is maximum when
sin2θ0 = 1 or θ0 =45°.
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• If the launch height and the
landing height are the same, then
the maximum horizontal range is
achieved if the launch angle is 45°.
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5/14
Projectile Motion Analyzed
• The launch height and the landing height differ.
• The launch angle 45° does not yield the
maximum horizontal distance.
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5/15
Projectile Motion Analyzed
The Effects of the Air
 Path I: Projectile movement if the air
resistance is taken into account
 Path II: Projectile movement if the air
resistance is neglected (as in a vacuum)
Our calculation along this chapter is based on
this assumption
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University Physics: Mechanics
5/16
Example: Baseball Pitcher
A pitcher throws a baseball at speed 40 km/h and at angle
θ = 30°.
h
30
(a) Determine the maximum height h
of the baseball above the ground.
v y  v 0 y  gt
0  5 .5 6  9 .8t
5.56
 0 .5 6 7 s
t
9.8
1
2
h  y  y 0  v 0 y t  2 gt
 (5.56)(0.567)  2 (9.8)(0.567)
 1.58 m
1
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4 0 k m h  1 1 .1 1 m s
v 0 y  v 0 sin 
 (11.11) sin 30 
 5.56 m s
v 0 x  v 0 co s 
 (1 1 .1 1) co s 3 0 
 9 .6 2 m s
x  x0  v0 xt
y  y0  v0 y t 
1
2
gt
2
v y  v 0 y  gt
2
v y  v0 y  2 g ( y  y 0 )
2
2
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5/17
Example: Baseball Pitcher
A pitcher throws a baseball at speed 40 km/h and at angle
θ = 30°.
30
d
(b) Determine the duration when the baseball is on the air.
t on air  t up  t dow n  0.567  0.567  1.134 s
(c) Determine the horizontal distance d
it travels.
d  x  x0  v0 x t
x  x0  v0 xt
y  y0  v0 y t 
 (9.62)(1.134)
 10.91 m
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1
2
gt
2
v y  v 0 y  gt
v y  v0 y  2 g ( y  y 0 )
2
2
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5/18
Example: Rescue Plane
A rescue plane flies at 198 km/h
and constant height h = 500 m
toward a point directly over a
victim, where a rescue capsule
is to land.
(a) What should be the angle Φ
of the pilot’s line of sight to
the victim when the capsule
release is made?
y  y 0  v0 y t 
1
2
1
gt
Released horizontally
2
d  x  x0
(  500)  (0)  (0) t  2 (9.8) t
2  500
2
t 
 102.041
9.8
t  1 0 .1 0 2 s
2
x  x0  v0 x t
 (55)(10.102)  5 5 5 .6 1 m
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h  y  y0
  tan
 tan
1
d
1
h
555.61
500
 48.016 
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5/19
Example: Rescue Plane
A rescue plane flies at 198 km/h
and constant height h = 500 m
toward a point directly over a
victim, where a rescue capsule
is to land.
(b) As the capsule reaches the
→
water, what is its velocity v
in unit-vector notation and in
magnitude-angle notation?
v y  v 0 y  gt
v y  (0)  (9.8)(10.102)
 99 m s
v x  v0 x
 55 m s
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Released horizontally
h  y  y0
d  x  x0
v  55iˆ  99 ˆj m s
Unit-vector notation
v  1 1 3 .2 5 2 m s   6 0 .9 4 5 
Magnitude-angle notation
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5/20
Example: Clever Stuntman
A stuntman plans a spectacular jump from
a higher building to a lower one, as can be
observed in the next figure.
Can he make the jump and safely reach
the lower building?
y  y 0  v0 y t 
1
2
1
gt
2
(  4.8)  (0)  (0) t  2 (9.8) t
2  4.8
2
t 
 0.98
9.8
t  0 .9 9 s
2
Time for the stuntman to fall 4.8 m
x  x0  v0 xt
x  x0  v0 x t
 (4.5)(0.99)
y  y0  v0 y t 
 4 .4 6 m
Horizontal distance jumped by the
stuntman in 0.99 s
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He cannot make
the jump
1
2
gt
2
v y  v 0 y  gt
v y  v0 y  2 g ( y  y 0 )
2
2
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5/21
Homework 5: Three Point Throw
New
A basketball player who is 2.00 m tall is standing on the floor
10.0 m from the basket. If he shoots the ball at a 40.0° angle
with the horizontal, at what initial speed must he throw so that it
goes through the hoop without striking the backboard? The
basket height is 3.05 m.
Erwin Sitompul
University Physics: Mechanics
5/22