introduction - city, university of london

3. Two-Dimensional Kinematics

1

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


2

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


3


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?


4


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


5


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.


6

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


7


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


8


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


9

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


10

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


11

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.


12

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


13

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


14

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?


15

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


16

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.


17

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


18

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?


19


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?


20


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.


21

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


22

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?


23

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°


24

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


25


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


26


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°


27

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?

introduction - city, university of london

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