Transcript Physics Definition

Physics
Two-Dimensional
Kinematics
Teacher: Luiz Izola
Chapter Preview
1. Motion in Two Dimensions
2. Projectile Motion Basic Equations
3. Zero Launch Angle
4. General Launch Angle
5. Projectile Motion Key Characteristics
Learning Objectives
 Apply the motion formulas in two
directions (x, y).
 Learn how to calculate motion on
projectiles
 Learn about circular motion.
Introduction
 This presentation will concentrate on the
study of motion in 2 dimensions.
 Particularly, we will concentrate on the
study of projectile motion and circular
motion.
 Projectile motion involves the study of
objects that are initially launched and
continue moving under gravity’s influence.
Force and Mass
 Chapter’s Main Idea:
“Horizontal and Vertical Motions
are independent.”
 For example: A ball thrown horizontally with a
speed v continues to move with the same speed v
in the horizontal direction, even as it falls with an
increasing speed in the vertical direction.
 The time of fall is the same whether a ball is
dropped from rest straight down or thrown
horizontally. Each motion continues independently
of the other.
Motion in Two Dimensions
 Motion with constant velocity, determining x an
y as functions of time.
Ex: A turtle starts at the origin at t=0 and moves with a
constant speed v0 = 0.26m/s as show below. How far
the turtle moved in the x, y directions after 5.0secs?
Motion in Two Dimensions
Ex2: An eagle perched on a tree limb 19.5m above the water
spots a fish swimming near the surface. The eagle pushes off
the branch and descends toward the water. By adjusting its
body in flight, the eagle maintains a constant speed of 3.10m/s
at an angle of 200 below the horizontal. (a) How long does it
take for the eagle to reach the water? (b) How long has the
eagle traveled in the horizontal when it reaches the water?
Constant Acceleration
To study motion with constant acceleration in two dimensions,
we repeat what was done with one dimension but we create two
sets of equations, one set for each dimension (x, y).
Table 4-1
Constant-Acceleration Equations of Motion
Position as a
function of time
Velocity as a function Velocity as a function
of time
of position
x = x0 + v0xt + ½ axt2
vx = v0x + axt
vx2 = v0x2 + 2axDx
y = y0 + v0yt + ½ ayt2
vy = v0y + ayt
vy2 = v0y2 + 2ayDx
Constant Acceleration
Ex: A hummingbird is flying in such a way that it is initially
moving vertically with a speed of 4.6m/s and accelerating
horizontally at 11m/’s2. Assuming the bird’s acceleration
remains constant for the time interval of interest, find the
horizontal and vertical distance through which it moves in
0.55s.
Projectile Motion Basic Equations

A projectile is an object that is thrown, kicked, batted, or
launched into motion and allowed to follow a path solely
determined by the influence of gravity.

In studying projectile motion, we make the following
assumptions:
I.
Air resistance is ignored
II.
The acceleration of gravity, g =9.8m/s2, and is constant
and downward.
III.
The Earth’s rotation is ignored.
Projectile Motion Basic Equations
 Suppose, in the figure below, that the x-axis is the
horizontal and the y-axis is the vertical, with the positive
direction upward. All objects in free fall have ax = 0 and
ay = -g. Since the downward direction is negative, it follows
that:
ay = 9.81m/s2 = -g
Projectile Motion Basic Equations
 Gravity cause no acceleration in the x direction.
ax = 0
 With these acceleration components placed into the
fundamental acceleration equations of motion, we have:
x = x0 + v0xt
vx = v0x
vx2 = v0x2
y = y0 + v0yt - ½gt2
vy = v0y -gt
vy2 = v0y2 - 2gDy
Zero Launch Angle

A special case is a projectile launched horizontally, so
that the angle between the initial velocity and the
horizontal is Θ = 0.
Equations of Motion:
 Choosing ground level to be y = 0 and the release point
to be directly above the origin, the initial position of the
ball is given by:
x0 = 0
y0 = 0
Zero Launch Angle


The initial velocity is horizontal, corresponding to Θ = 0,
see figure next slide. As a result, the x component of the
initial velocity is simply the initial speed,
v0x = v0cos(Θ) = v0
And the y component:
v0y = v0sin(Θ) = 0
Replacing these values in the motion formulas we have:
x = v0t
vx = v0
vx2 = v02
y = h - ½gt2
vy = -gt
vy2 = - 2gDy
Zero Launch Angle
Zero Launch Angle
Ex: A person walking with a speed of 1.30m/s releases a ball
from a height of 1.25m above the ground. Given that x0 = 0
and y0 = h = 1.25m, find x and y for (a) t = 0.250s and
(b) t = 0.50s. (c) Find the velocity, speed, and direction of the
ball at t = 0.50s. (d) How long does it take for the ball to land?
Parabolic Path and Landing Site
 The curved path of the projectile can be found by
combining the following:
x = v0t
and
h = -gt2
 This will give:
y = h – 1/2g(x/v0)x2
 Where does a projectile land if it is launched horizontally
with speed v0 from a height h?
By setting y = 0 in the above equation, we have:
x = v0 (2h/g)
Parabolic Path and Landing Site
Ex: A mountain climber encounters a crevasse in an ice field.
The opposite side of the crevasse is 2.75m lower, and is
separated horizontally by a distance of 4.10m. To cross the
crevasse, the climber gets a running start and jumps in the
horizontal direction. (a) What is the minimum speed
needed by the climber to safely cross the crevasse? If,
instead, the climber’speed is 6.00m/s, (b) where does the
climber land?, and (c) What is the climber’s speed at
landing?
Practice Problems
1. A sailboat runs before the wind with a constant
speed of 2.8m/s in a direction of 520 north of
west. How far (a) west and (b) north has the
sailboat traveled in 35min?
2. Starting from rest a car accelerates at 2.0m/s2
up a hill that is inclined 5.50 above the
horizontal. How far (a) horizontally, and (b)
vertically has the car traveled in 12s?
General Launch Angle

Consider now, the case of a projectile launched at an
arbitrary angle with respect to the horizontal. To simplify the
resulting equation, we take the launching site to be the
origin.
General Launch Angle
Since the projectile starts at the origin, we have:
x0 = y0 = 0
The components of the initial velocity are:
v0x = v0cos(Θ)
v0y = v0sin(Θ)
The resulting equations are:
x = v0cos(Θ)t
vx = v0cos(Θ)
vx2 = v02cos2(Θ)
y = v0sin(Θ) t - ½ gt2
vy = v0sin(Θ) - gt
vy2 = v02sin2(Θ) – 2gDy
Practice Problems
1.
Chipping from the rough a, a golfer sends the ball over a
3.00m-high tree that is 14.0m away. The ball lands at the
same level from which it was struck after traveling an
horizontal distance of 17.8m. (a) If the ball left the club 540
above the horizontal and landed on the green 2.24s later,
what was its initial speed? (b) How high was the ball when
it passed over the tree?
Practice Problems
2.
A projectile is launched from the origin with an initial speed
of 20.0m/s at an angel of 350 above the horizontal. Find the
x,y positions of the projectile at times (a) t=0.50s, (b)
t=1.00s, and (c) t = 1.50s.
3.
A golfer hits a ball with an initial speed of 30m/s at an angle
of 500 above the horizontal. The ball lands on a green that
is 5.00m above the level where the ball was struck.
a)
b)
How long is the ball in the air?
How far has the ball traveled in the horizontal when it
lands?
What is the speed and direction of motion of the ball just
before it lands?
c)
Practice Problems
A trained dolphin leaps from the water with an initial speed of
12m/s (see figure below). In the absence of gravity, the
dolphin would move on a straight line to the ball and catch
it, but because of gravity, the dolphin follows a parabolic
path well below the ball’s initial position. If the trainer
releases the ball the instant the dolphin leaves the water,
show where the dolphin and falling ball meet.
Projectile Motion: Key Characteristics
RANGE
The range, R, of a projectile is the horizontal distance it travels
before landing. Considering initial and final elevations
the same (y = 0).
Challenge: Using the assumption about y = 0 above and the
following formulas of motion:
y = v0sin(Θ) t - ½ gt2
x = v0cos(Θ)t
Prove that R is equals to:
R = (v02/g ).sin2Θ
Practice Problems
1)
A football game begins with a kickoff in which the ball
travels a horizontal distance of 41m. If the ball is kicked at
an angle 400 above the horizontal, what was its initial
speed?
2)
A golf ball is struck on level ground. It lands 92.2m away
4.30s later. What was the direction and magnitude of the
initial velocity?
3)
The men’s world’s record for the shot put, 23.12m, was set
by Randy Barnes of the USA on May 20, 1990. If the shot
was launched from 1.80m above the ground at an initial
angle of 420, what was the initial speed?
Maximum Height
Challenge: Using the two formulas below
y = v0sin(Θ) t - ½ gt2
vy = v0y - gt
Prove that the maximum height can be calculate as:
h = v0sin(Θ) / 2g
Practice Problems
The archerfish hunts by dislodging an unsuspecting insect from
its resting place with a stream of water expelled from the
fish’s mouth. Suppose the archerfish squirts water with an
initial speed of 2.30m/s at an angle of 19.50 above the
horizontal. When the stream of water reaches a beetle on the
leaf at height h above the water’s surface it is moving
horizontally. (a) How much time does the beetle have to
react? (b) What is the beetle’s height? (c) What the horizontal
distance between the beetle and the fish?
Homework
I.
Page 102: 11, 13, 15, 17
II.
Page 103: 19, 21, 23
III.
Page 104: 25

Note: It would be excellent to try all the projectile motion
problems in order to get more proficient.