INTRODUCTION - City, University of London
Transcript of INTRODUCTION - City, University of London
3. Two-Dimensional Kinematics
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INTRODUCTION
• We now extend our study of kinematics to motion in two
dimensions (x and y axes)
• This will help in the study of such phenomena as
projectile motion
• Projectile motion is the study of objects that are initially
launched (or projected) and then continue moving
under the influence of gravity alone
• We will see that 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
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MOTION IN 2D: CONSTANT
VELOCITY
• Consider the situation shown above, where a turtle
starts from the origin at t = 0, and moves with constant
speed v0 = 0.26m/s, 25°above the x axis
• In 5.0s, the turtle travels d = v0t = 0.26×5 = 1.3m in a straight line
• In the x direction: x = d cos(25) = 1.2m
• In the y direction: y = d sin(25) = 0.55m
• Alternatively, we could treat the x and y motions
separately
• The x velocity component is: v0x = v0 cos(25) = 0.24m/s
• The y velocity component is: v0y = v0 sin(25) = 0.11m/s
• Displacement x is: x = v0xt = 1.2m
• Displacement y is: y = v0yt = 0.55m
• Generally the turtle might start at a position x = x0 at
time t = 0
• Thus: x = x0 + v0xt and y = y0 + v0yt
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MOTION IN 2D: CONSTANT
VELOCITY - EXAMPLE
• An eagle perched on a tree branch 19.5m above the
water spots a fish swimming near the surface. The
eagle pushes off from the branch and descend toward
the water. By adjusting its body in flight, the eagle
maintains a constant speed of 3.1m/s at an angle of 20°
below the horizontal. How long does it take for the
eagle to reach the water? How far has the eagle
traveled in the horizontal direction when it reaches the
water?
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MOTION IN 2D: CONSTANT
ACCELERATION
• In 1D, we used x = x0 + v0t + ½ at2
• In 2D, we replace both v0 and a with the corresponding
x components, v0x and ax
• Thus x = x0 + v0xt + ½ at2
• For the y direction: y = y0 + v0yt + ½ at2
• The above are position versus time equations of motion
for two dimensions
• The same approach gives velocity as a function of time
• Start with v = v0 + at and write it in terms of x and y
components
• Thus: vx = v0x + axt and vy = v0y + ayt
• We can write v2 = v02 + 2a∆x in terms of components
• Thus: vx2 = v0x
2 + 2ax∆x and vy2 = v0y
2 + 2ay∆y
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MOTION IN 2D: CONSTANT
ACCELERATION - EXAMPLE
• 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.
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PROJECTILE MOTION: BASIC
EQUATIONS (1)
• Here we will consider the independence of horizontal
and vertical motions to projectiles
• A projectile is an object that is thrown, kicked, batted, or
otherwise launched into motion and then allowed to
follow a path determined solely by gravity
• In studying projectile motion, the following assumptions
are made
– Air resistance is ignored
– The acceleration due to gravity is constant, downward and
has a magnitude g = 9.81m/s2
– The Earth’s rotation is ignored
• Air resistance can be significant when a projectile
moves with relatively high speed or in strong winds
• The value of g varies slightly from place to place on the
Earth’s surface and decreases with altitude
• The rotation of the Earth can be significant when
considering projectiles that cover great distances
• Yet for the simple situation of dropping a ball, such
drawbacks can be ignored
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PROJECTILE MOTION: BASIC
EQUATIONS (2)
• Consider the above figure, where the x axis is
horizontal and the y axis is vertical, with upwards being
the positive direction
• Since downwards is negative: ay = -9.81m/s2 = -g
• Gravity causes no acceleration in the x direction, and
so ax = 0
• So for projectile motion we have the following:
• x = x0 + v0xt vx = v0x vx2 = v0x
2
• y = y0 + v0yt - ½ gt2 vy = v0y - gt vy2 = v0y
2 - 2g∆y
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PROJECTILE MOTION: BASIC
EQUATIONS - DEMONSTRATION
• A simple demonstration illustrates the independence of horizontal and vertical motions in projectile motion
• Firstly while standing still, a rubber ball is dropped and caught on the rebound from the floor
• The ball goes straight down, lands near your feet and returns almost to the level of your hand
• Next walk or skate with constant speed and drop the ball, and observe its motion
• To you, the motion looks the same as when standing still, i.e. it lands near your feet and bounces straight back up
• The fact you were moving horizontally had no effect on the ball’s vertical motion – motions were independent
• To an observer who sees you walking by, the ball follows a curved path as shown
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LAUNCH ANGLE OF A PROJECTILE
• A projectile launched at
an angle above the
horizontal where θ > 0
• A launch below the
horizontal would
correspond to θ < 0
• A projectile launched
horizontally is when θ =
0
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ZERO LAUNCH ANGLE: EQUATIONS
OF MOTION
• A special case is a projectile is launched horizontally so
that the angle between the initial velocity and the
horizontal is θ = 0
• Taking into consideration the figure on the previous
slide, if we choose the ground level to be y = 0, with
height h and walking speed v0, we can say that the
initial position of the ball is given by x0 = 0 and y0 = h
• The initial velocity is horizontal, which corresponds to θ
= 0, and as a result the x and y velocity components
are: v0x = v0 cos(0) = v0 and v0y = v0 sin(0) = 0
• Substituting these specific values into the fundamental
equations for projectile motion yield:
• x = v0t vx = v0 = constant vx2 = v0
2 = constant
• y = h - ½ gt2 vy = - gt vy2 = v0y
2 - 2g∆y
• Note that the x component of velocity remains the same
for all time and that the y component steadily increases
with time
• As a result, x increases linearly with time, and y
decreases with t2 dependence
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ZERO LAUNCH ANGLE: EXAMPLE
• A person skateboarding with a constant speed of
1.3m/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 t = 0.25s and t = 0.5s. Find the velocity, speed and
direction of motion of the ball at t = 0.5s.
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PARABOLIC PATH
• Just what is the shape of the curved path followed by a
projectile launched horizontally?
• This is found by combining x = v0t and y = h - ½ gt2
• Combining these equations allows us to express y in
terms of x by eliminating t, i.e. t = x/v0
• Substituting this result into the y equation to eliminate t
yields
• y = h – ½ g(x/v0)2 = h – (g/2v0
2)x2
• This has the form y = a + bx2, where a = h and is
constant
• Also b = – g/2v02 and is also constant
• This is the equation of a parabola that curves
downwards, a characteristic shape in projectile motion
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LANDING SITE
• Where does a projectile land if it is launched
horizontally with a speed v0 from a height h?
• The most direct way to solve this is to set y = 0, since y
= 0 corresponds to ground level
• Thus 0 = h – (g/2v02)x2
• Solving the above equation for x yields the landing site
and thus x = v0√(2h/g)
• Note that the positive sign for the square root has been
chosen since the projectile was launched in the positive
x direction, and hence lands at a positive value of x
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PARABOLIC PATH AND LANDING
SITE EXAMPLE
• A mountain climber encounters a crevasse in an ice
field. The opposite size of the crevasse is 2.75m lower
and is separated horizontally by a distance of 4.1m. To
cross the crevasse, the climber gets a running start and
jumps in the horizontal direction. What is the minimum
speed needed by the climber to safely cross the
distance? If, instead, the climber’s speed is 6.0m/s,
where does the climber land and what is the climber’s
speed on landing?
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GENERAL LAUNCH ANGLE
• The more general case of
a projectile being launched
from an arbitrary angle with
respect to the horizontal
will be considered
• The simplifications made
with zero launch angle are
no longer used
• Here θ is nonzero
• For a projectile launched
with initial speed v0 at
angle θ, the initial x and y
positions are zero, as it
starts at the origin, thus x0
= y0 = 0
• The middle figure shows
the components of the
initial velocity
• The last figure shows the
conditions when θ = 90°
and 0
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GENERAL LAUNCH ANGLE
EQUATIONS
• Substituting the results
from the previous slide
yield the following
• x = (v0cosθ)t
• vx = v0cosθ
• vx2 = v0
2cos2θ
• y = (v0sinθ)t – ½ gt2
• vy = v0sinθ – gt
• vy2 = v0
2sin2θ – 2g∆y
• These equations are valid
for any launch angle
• They reduce to the simpler
equations derived earlier
when θ = 0 and y0 = h
• Example: A projectile is
launched from the origin
with initial speed 20.0m/s
at an angle of 35°above
the horizontal. Find the x
and y positions at times t =
0.5s, t = 1.0s and t = 1.5s.
Also find the velocity at
these times.
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SNAPSHOTS OF A TRAJECTORY
• Above shows the projectile referred to in the previous
example
• The points are not evenly spaced in terms of position
even though they are spaced evenly in time
• The points bunch closer together at the top of the
trajectory which indicates that a comparatively large
fraction of the flight time is spent near the highest point
• This gives the illusion that an object hangs in the air,
e.g. a basketball player performing a slam dunk
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GENERAL LAUNCH ANGLE:
EXAMPLE 1
• Chipping from the rough, a golfer sends the ball over a
3.0m high tree that is 14.0m away. The ball lands at the
same level from which it was struck after travelling a
horizontal distance of 17.8m on the green. If the ball left
the club 54°above the horizontal and landed on the
green 2.24s later, what was the initial speed? How high
was the ball when it passed over the tree?
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GENERAL LAUNCH ANGLE:
EXAMPLE 2
• A golfer hits a ball from the origin with an initial speed
of 30.0m/s at an angle of 50°above the level where the
ball was struck.
• How long is the ball in the air?
• How far has the ball travelled in the horizontal direction
when it lands?
• What is the speed and direction of motion of the ball
just before it lands?
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GENERAL LAUNCH ANGLE:
EXAMPLE 3
• A trained dolphin leaps from the water with an initial
speed of 12.0m/s. It jumps directly toward a ball held by
a trainer a horizontal distance of 5.5m away and a
vertical distance of 4.1m above the water. In the
absence of gravity the dolphin would move in 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, as shown. If the trainer releases the ball
the instant the dolphin leaves the water, show that the
dolphin and the falling ball meet.
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PROJECTILE MOTION: RANGE
• The range, R, is the horizontal distance a projectile travels before landing
• Consider the above where the initial and final elevations are the same (y = 0)
• To find the range use y = (v0sinθ)t – ½ gt2 to find tfor y = 0 and substitute this time into the xequation of motion
• Thus for y = 0: (v0sinθ)t = ½ gt2 so (v0sinθ) = ½ gt
• Therefore t = (2v0/g)sinθ (time of flight)
• Substitute this into x = (v0cosθ)t which yields
• x = (2v02/g)sinθcosθ, and x = R, sin2θ = 2sinθcosθ
• Thus the range R = (v02/g)sin2θ
• Above equation only valid for same initial and final elevation
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RANGE: EXAMPLE
• A rugby game begins with a kickoff in which the ball
travels a horizontal distance of 41m and lands on the
ground. If the ball was kicked at an angle of 40°above
the horizontal, what was its initial speed?
• Suppose the initial speed of the ball is increased by
10%. By what percentage does the range increase?
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MAXIMUM RANGE
• R depends inversely on g, so the smaller g the larger
the range
• On the moon, the acceleration due to gravity is about
1/6 of g, so a projectile would travel 6 times as far as it
would on Earth
• What launch angle gives the greatest range?
• We know that R varies with sin2θ and so R is a
maximum when sin2θ is a maximum, i.e. when sin2θ =
1 which is when sin(90) = 1, thus θ = 45°
• Since the maximum range occurs when sin2θ = 1
• Rmax = (v02/g) – only valid when the projectile lands at
the same level from which it was launched
• Also it is only valid for the ideal case of no air
resistance
• Air resistances come into consideration for fast moving
objects, and the range is reduced
• The maximum range occurs for launch angles less that
45°
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SYMMETRY IN PROJECTILE
MOTION (1)
• Recall that the time when the projectile lands is given
by t = (2v0/g)sinθ
• By symmetry the time it takes a projectile to reach its
highest point (in the absence of air resistance) should
be half this time
• After all, the projectile moves in the x direction with
constant speed, and the highest point (i.e. the
maximum y) occurs at x = ½ R
• To prove this, it is safe to say that at the highest point
the projectile is moving horizontally, so vy = 0
• We would like to find the time when vy = 0
• Thus vy = voy – gt = v0sinθ – gt = 0
• Therefore t = (v0/g)sinθ, which is half the time before
the projectile lands
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SYMMETRY IN PROJECTILE
MOTION (2)
• There is another interesting symmetry for speed
• Recall that when a projectile is launched vy = v0sinθ
• When the projectile lands at t = (2v0/g)sinθ, vy is
• vy = v0sinθ – gt = v0sinθ - g(2v0/g)sinθ = -v0sinθ
• This is the exact opposite of the vy when it was
launched
• Since vx is always the same, it follows that when the
projectile lands, its speed v = √(vx2 + vy
2) and is the
same as when it was launched
• But the velocities are different since the directions of
motion are different at launch and landing
• The symmetry described here extends to any level
• At a given height the speed of a projectile is the same
on the way up as on the way down
• The angle of the velocity above the horizontal on the
way up is same as the angle below the horizontal on
the way down, as shown
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SYMMETRY IN PROJECTILE
MOTION (3)
• Consider the range R – above left shows R versus
launch angle θ for v0 = 20m/s
• As discussed, without air resistance, R is a maximum
when θ is 45°
• Note from the plot that the range for angles equally
above or below 45°is the same, i.e. the range for 30°is
the same for 60°
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MAXIMUM RANGE
• Recall that a projectile is at a maximum height when
the y component of velocity is zero
• This can be used to determine the maximum height of
an arbitrary projectile
• First, find the time when vy = 0
• Then substitute this time into the y versus t equation
where y = (v0sinθ)t – ½ gt2
• Example: 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 2.3m/s
at an angle of 19.5°above the horizontal. When the
stream of water reaches a beetle on a leaf at height h
above the water’s surface, it is moving horizontally.
How much time does the beetle have to react? What is
the height h of the beetle? What is the horizontal
distance d between the fish and the beetle when the
water is launched?