Transcript Slide 1
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
then 966 km south from city
B
to city to city
C B
in 45 min and 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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B C
Δr 2 Δr 1
Solution of Homework 4: The Plane
483 km, 45 min
A
r
total
t
1 2 total
t
1 1
t r
2 2 1
t
2 45 min 1.5 h 483i ˆ 0.75 h 2.25 h
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(a) the magnitude of its displacement
r
total 2 1080.021 km 2 243.435
B A
(b) the direction of its displacement
r
total tan 1 966 483 63.435
243.435
• •
Quadrant I Quadrant III
C
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Solution of Homework 4: The Plane
(c) the magnitude of its average velocity
v
avg
r
total
t
total 483i ˆ 2.25 h
v
avg 2 2 480 km h (d) the direction of its average velocity
v
avg tan 1 429.333
214.667
243.435
•
Quadrant III
(e) its average speed
s
avg total distance traveled total time 2.25 h 644 km h
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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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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.
a x
0
g
9.81m s
2
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a y
g
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Projectile Motion
a x
0
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v
0
v
0
x v
0
y
v
0
v
0
v
0
x
ˆ i
cos
sin
v
0 0 0
y
ˆ j
a y
g
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Two Golf Balls
v
0
x
0
v
0
x
0
Projectile Motion
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• •
The vertical motions are quasi identical.
The horizontal motions are different.
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Projectile Motion Analyzed
The Horizontal Motion
x
x
0
v t
0
x x
x
0 (
v
0 cos 0 )
t
The Vertical Motion
y
y
0
v t
0
y
1 2
gt
2
v
2
y y
y
0
v y v
0 (
v
0
v
0 0 sin sin ) 2 0 0 )
t
gt
1 2
gt
2
y
0 )
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v v x y
= v 0x = –v 0y Erwin Sitompul
Projectile Motion Analyzed
The Horizontal Range
R x
(
v
0
x
0 cos
R
0 )
t
0 (
v
0
y
sin
y
0 0 )
t
0 1 2
gt
2 Eliminating
t
,
R
2
v
0 2
g
sin 0 cos 0 •
This equation is valid if the landing height is identical with the launch height.
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Projectile Motion Analyzed
Further examining the equation, 2
R
2
v g
0 sin 0 cos 0 Using the identity sin 2 0 2sin 0 cos 0 , we obtain 2
R
v g
0 sin 2 0
R
is maximum when sin2
θ
0 = 1 or
θ
0 =45 °. •
If the launch height and the landing height are the same, then the maximum horizontal range is achieved if the launch angle is 45 °. Erwin Sitompul University Physics: Mechanics 5/11
Symmetry of Position and Speed
v
0 = 29.4 m/s • •
If the initial elevation and final elevation are the same, the velocity of an object at each elevation will be the same in magnitude , but opposite in direction.
The object’s height and the speed will be symmetrical around the time when the peak position is reached.
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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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The Effects of the Air
Projectile Motion Analyzed
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 Erwin Sitompul University Physics: Mechanics 5/14
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 5.56 9.8
t h t
5.56
9.8
y
0 1.58 m 0.567 s
v t
0
y
(5.56)(0.567) 2 1 1 2
gt
2 (9.8)(0.567) 2
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v
0
y v
0
x
v
0 sin (11.11) sin 30 5.56 m s
v
0 cos (11.11) cos 30 9.62 m s
v
2
y y
v x y
0
x
0
v t
0
y v t
0
x
1 2
y
v
0
y
gt v
2 0
y
gt y
2 0 )
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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
down 1.134 s (c) Determine the horizontal distance
d
it travels.
d x x
0
v t
0
x
(9.62)(1.134) 10.91 m
v
2
y y
v x y
0
x
0
v t
0
y v t
0
x
1 2
y
v
0
y
gt v
2 0
y
gt y
2 0 )
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Simulation: How to Fire the Cannon?
θ v
0 9 m 16 m
A cannon is 1.20 m above the ground. You may adjust the initial speed and the angle of fire of the cannon.
If the target is horizontally 16 m away from the cannon and at 9 m above the ground, how do you set the cannon so that the projectile can hit the target?
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v
2
y y
v x y
0
x
0
v t
0
y v t
0
x
1 2
y
v
0
y
gt v
2 0
y
gt y
2 0 )
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Example: Rescue Plane
A rescue plane flies at 198 km/h and constant height is to land.
h
= 500 m toward a point directly over a victim, where a rescue capsule (a) What should be the angle release is made?
Φ
of the pilot’s line of sight to the victim when the capsule
y
y
0
v t
0 (0)
y t
1 2 1 2
gt
2 (9.8)
t
2
t
2
t
9.8
10.102 s 102.041
x
x
0
v t
0
x
(55)(10.102) 555.61 m
Released horizontally
h
d y
0
x
0 1
d
tan 1
h
555.61
tan 48.016
500
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Example: Rescue Plane
Released horizontally
A rescue plane flies at 198 km/h and constant height is to land.
h
= 500 m toward a point directly over a victim, where a rescue capsule (b) As the capsule reaches the water, what is its velocity →
v
in unit-vector notation and in magnitude-angle notation?
v y v y
v
0
y
gt
99 m s
v x
v
0
x
55 m s
h d y
0
x
0
v
ˆ
Unit-vector notation
v
113.252 m s 60.945
Magnitude-angle notation Erwin Sitompul University Physics: Mechanics 5/19
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 1
v t
0
y
(0)
t
2 1 2
gt
2 (9.8)
t
2
t
2
t
0.98
9.8
0.99 s
Time for the stuntman to fall 4.8 m
x
x
0
v t
0
x
(4.5)(0.99) 4.46 m
Horizontal distance jumped by the stuntman in 0.99 s Erwin Sitompul He cannot make the jump
v
2
y y
v x y
0
x
0
v t
0
y v t
0
x
1 2
y
v
0
y
gt v
2 0
y
gt y
2 0 )
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Homework 5A: Three Point Throw
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.
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Homework 5B: Docking the Ship
1. A dart player throws a dart horizontally at 12.4 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board?
2. As a ship is approaching the dock at 45.0 cm/s, an important piece of landing equipment needs to be thrown to it before it can dock. This equipment is thrown at 15.0 m/s at 60.0
° above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship’s deck.
For this equipment to land at the front of the ship, at what distance
D
from the dock should the ship be when the equipment is thrown? Air resistance can be neglected.
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