Random Physics articles

8/11/2019 Random Physics articles

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The Planck Hypothesis

In order to explain the frequency distribution of radiation from a hot cavity

( blackbody radiation) Planck proposed the ad hoc assumption that the radiant energy

could exist only in discrete quanta which were proportional to the frequency. Thiswould imply that higher modes would be less populated and avoid the ultraviolet

catastrophe of the Rayleigh-Jeans Law.

The quantum idea was soon seized to explain the photoelectric effect, became part of

the Bohr theory of discrete atomic spectra, and quickly became part of the foundation

of modern quantum theory.

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The Photoelectric Effect

The details of the photoelectric effect were indirect contradiction to the expectations of verywell developed classical physics.

The explanation marked one of the major stepstoward quantum theory.

The remarkable aspects of the photoelectric

effect when it was first observed were:

1. The electrons were emitted immediately -no time lag!

2. Increasing the intensity of the light

increased the number of photoelectrons, but

not their maximum kinetic energy!

3. Red light will not cause the ejection of

electrons, no matter what the intensity!4. A weak violet light will eject only a few

electrons, but their maximum kinetic

energies are greater than those for intenselight of longer wavelengths!















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Properties of Molecules

The investigation of molecular structure parallels the study of atomic structure in that

the methods of quantum mechanics are applied along with the information obtained

from molecular spectra.

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Molecular Spectra



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The most commonly observed molecular

spectra involve electronic, vibrational, or

rotational transitions. For a diatomicmolecule, the electronic states can be

represented by plots of potential energy as a

function of internuclear distance. Electronictransitions are vertical or almost vertical lines

on such a plot since the electronic transition

occurs so rapidly that the internucleardistance can't change much in the process.

Vibrational transitions occur between

different vibrational levels of the same

electronic state. Rotational transitions occurmostly between rotational levels of the same

vibrational state, although there are many

examples of combination vibration-rotation

transitions for light molecules.http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c1

Rotational Spectra

Incident electromagnetic waves can excite therotational levels of molecules provided they

have an electric dipole moment. The

electromagnetic field exerts a torque on themolecule. The spectra for rotational transitionsof molecules is typically in

the microwaver egion of the electromagnetic

spectrum. The rotational energies for rigidmolecules can be found with the aid of the

Shrodinger equation. The diatomic molecule can

serve as an example of how the determined

moments of inertia can be used to calculate bond lengths.

The illustration at left shows some perspectiveabout the nature of rotational transitions. The

diagram shows a portion of the potential

diagram for a stable electronic state of adiatomic molecule. That electronic state will

have several vibrational states associated with it,

so that vibrational spectra can be observed.Most commonly, rotational transitions which are



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associated with the ground vibrational state are

observed.

Rotational Energies

The classical energy of a freely rotating molecule can be expressedas rotational kinetic energy

where x, y, and z are the principal axes of rotation and Ix representsthe moment of inertia about the x-axis, etc. In terms of the angular

momenta about the principal axes, the expression becomes

The formation of the Hamiltonian for a freely rotating molecule isaccomplished by simply replacing the angular momenta with the

corresponding quantum mechanical operators.

Diatomic molecules

Index

Molecular

spectra

concepts

Go Back

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HyperPhysics***** Quantum Physics R

Nave

Diatomic Molecules

For a diatomic molecule the rotational energy is obtained fromthe Schrodinger equation with the Hamiltonian expressed in terms of the

angular momentum operator.

More detail

where J is the rotational angular momentum quantum number and I isthe moment of inertia.

Rotational transitions

Determining the rotational constant B

enables you to calculate the bond lengthR. The allowed transitions for thediatomic molecule are regularly spaced at

interval 2B. The measurement and

identification of one spectral line allows

one to calculate the moment of inertia and

then the bond length.Examples HCl CN CH

Table of diatomic molecule parameters

Index

Molecular

spectra

concepts


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Rotational Transitions, Diatomic

For a rigid rotor diatomic molecule, the selection rules for rotational

transitions are ΔJ = +/-1, ΔMJ = 0 .

The rotational spectrum of a diatomicmolecule consists of a series of

equally spaced absorption lines,

typically in themicrowave region ofthe electromagnetic spectrum. Theenergies of the spectral lines are

2(J+1)B for the transitions J -> J+1.

For real molecules like the example of

HCl , the successive transitions are a

bit lower than predicted becausecentrifugal distortion lengthens the

molecule, increasing its moment of

inertia.

Moment of inertia for diatomic molecule

Vibration/rotation transitions in HCl

Index

Molecularspectra

concepts


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The following is a sampling of transition

frequencies from the n=0 to n=1 vibrational

level for diatomic molecules and thecalculated force constants.

Molecule Frequencyx10

13 Hz

Force constant N/m

HF 12.4* 970

HCl 8.66 480

HBr 7.68 410

HI 6.69 320

CO 6.42 1860

NO 5.63 1530

* From vibrational transition 4138.52 cm-1 in Herzberg'stabulation.

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Vibration-Rotation Spectrum of HCl

Add annotation to spectrum

A classic among molecular spectra, the infrared absorption spectrum of HCl

can be analyzed to gain information about both rotation and vibration of the

molecule.

The absorption lines shown involve transitions from the ground to first excited

Index

Molecular

spectra

concepts

Reference

Tipler &

Llewellyn Sec. 9-4

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vibrational state of HCl, but also involve changes in the rotational state. Therotational angular momentum changes by 1 during such transitions. If you hada transition from j=0 in the ground vibrational state to j=0 in the first excited

state, it would produce a line at the vibrational transition energy. As observed,

you get a closely spaced series of lines going upward and downward from that

vibrational level difference. The splitting of the lines shows the difference inrotational inertia of the two chlorine isotopes Cl-35(75.5%) and Cl-37(24.5%).

From the spectrum above, you can examine details about the following:

Bond force constant Bond length

j for peak intensity Relative intensities

Pure rotational transitions in HCl

Vibration-rotation spectrum of HBr

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Bond Force Constant for HCl

By treating the vibrational transition in the HClspectrum from its ground to first excited state as a quantum

harmonic oscillator , the bond force constant can becalculated. This transition frequency is related to the

molecular parameters by:

The desired transition frequency does not show up directlyin the observed spectrum, because there is no j=0, v=0 to

j=0, v=1 transition; the rotational quantum number must

change by one unit. It can be approximated by the midpoint between the j=1,v=0->j=0,v=1 transition and the j=0,v=0-

>j=1,v=1 transition. This assumes that the difference

between the j=0 and j=1 levels is the same for the groundand first excited state, which amounts to assuming that the

first excited vibrational state does not stretch the bond. This

"rigid-rotor" model can't be exactly correct, so it introducessome error.

For the HCl molecule, the needed reduced mass is

Note that this is almost just the mass of the hydrogen. Thechlorine is so massive that it moves very little while the

hydrogen bounces back and forth like a ball on a rubber

band!

Substituting the midpoint frequency into the expression

containing the bond force constant gives:

Index

Molecula

r spectraconcepts

Reference

Tipler &

Llewelly

n Sec. 9-4





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Despite the approximations, this value is quite close to the value given in the table.

Pure rotational transitions in HCl

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Bond Length of HCl















A bond length for the HCl molecule can be calculated from

the HCl spectrum by assuming that it is a rigid rotor andsolving the Schrodinger equation for that rotor. For a free

diatomic molecule the Hamiltonian can be anticipated from

the classical rotational kinetic energy

and the energy eigenvalues can be anticipated from the nature

of angular momentum.

Assuming that the bond length is the same for the ground andfirst excited states, the difference between the j=1,v=0-

>j=0,v=1 transition and the j=0,v=0->j=1,v=1 transitionfrequencies can be used to estimate the bond length. The

separation between the two illustrated vibration-rotation

transitions is assumed to be twice the rotational energychange from j=0 to j=1.

Substitution of numerical values leads to an estimate of the bond length r:

This compares reasonably with the value r=.127 nm obtained from pure rotational

spectra.

















Other Heteronuclear Diatomic

Molecules

Molecule Dissociation Energy(eV)Equilibrium Separation (nm)

(Bond length)

BN 4.0 0.128

CO 11.2 0.113

HBr ... 0.141

HCl 4.4 0.127

HF 5.8 0.092

NO 7.0 0.115

PbO 4.1 0.192

PbS 3.3 0.239

Ionic Diatomic Molecules Homonuclear Diatomic Molecules

Index

Tables

Data

References

Krane Ch 9

Beiser,

Concepts... Sec 13.5


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Properties of Ionic Diatomic Molecules


(Bond length)

NaCl 4.26 0.236 NaF 4.99 0.193

NaBr 3.8 0.250

NaI 3.1 0.271

NaH 2.08 0.189

LiCl 4.86 0.202







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LiH 2.47 0.239

LiI 3.67 0.238

KCl 4.43 0.267

KBr 3.97 0.282

RbF 5.12 0.227

RbCl 4.64 0.279

CsI 3.57 0.337


Homonuclear Diatomic Molecules


(Bond length)

H-H 4.5 0.075

N-N 9.8 0.11

O-O 5.2 0.12

F-F 1.6 0.14

Cl-Cl 2.5 0.20










Raman Scattering

When light encounters molecules in the air, the predominant mode of

scattering is elastic scattering, called Rayleigh scattering. This scattering

is responsible for the blue color of the sky; it increases with the fourth power of the frequency and is more effective at short wavelengths. It is

also possible for the incident photons to interact with the molecules insuch a way that energy is either gained or lost so that the scattered photons

are shifted in frequency. Such inelastic scattering is called Raman

scattering.

Like Rayleigh scattering, the Raman scattering depends upon the

polarizability of the molecules. For polarizable molecules, the incident

photon energy can excite vibrational modes of the molecules, yieldingscattered photons which are diminished in energy by the amount of the

vibrational transition energies. A spectral analysis of the scattered lightunder these circumstances will reveal spectral satellite lines below theRayleigh scattering peak at the incident frequency. Such lines are called

"Stokes lines". If there is significant excitation of vibrational excited states

of the scattering molecules, then it is also possible to observe scattering atfrequencies above the incident frequency as the vibrational energy is

added to the incident photon energy. These lines, generally weaker, are

called anti-Stokes lines.

Although finding some application in vibrational spectroscopy of

molecules, the use of direct infrared sources for such spectroscopy is

usually much easier. Raman spectroscopy has found some application inremote monitoring for pollutants. For example, the scattering produced by

a laser beam directed on the plume from an industrial smokestack can be

used to monitor the effluent for levels of molecules which will producerecognizable Raman lines.

Raman scattering can also involve rotational transitions of the moleculesfrom which the scattering occurs. Thornton and Rex picture a photon of

energy slightly than the energy separation of two levels being scattered,

with the excess energy released in the form of a photon of lower energy.

Since this is a two-photon process, the selection rule is J = +/-2 for

rotational Raman transitions. The sketch below is an idealized depiction ofa Raman line produced by interaction of a photon with a diatomic

molecule for which the rotational energy levels depend upon one moment

of inertia. The upper electronic state of such a molecule can have differentlevels of rotational and vibrational energy. In this case the upper state is

shown as being in rotational state J with scattering associated with an

incoming photon at energy matching the J+2 state.

Index

Scattering

concepts

Atmospheric

opticsconcepts

Molecular

spectraconcepts

ReferenceThornton

and Rex

Sec 11.1

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Since the Raman effect depends upon the polarizability of the molecule, itcan be observed for molecules which have no net dipole moment andtherefore produce no pure rotational spectrum. This process can yield

information about the moment of inertia and hence the structure of the

molecule.

In Raman scattering, an intense monochromatic light source (laser) can

give scattered light which includes one or more "sidebands" that are offset by rotational and/or vibrational energy differences. This is potentially very

useful for remote sensing, since the sideband frequencies contain

information about the scattering medium which could be useful foridentification. Current projects envision Raman scattering as a tool for

identification of mineral forms on Mars. Such remote sensing could

become a major tool in planetary exploration.

Raman scattering from lunar soil

Some history

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C. V. Raman

C. V. Raman discovered the inelastic scattering phenomenon which bears his name in1928 and for it he was awarded the Nobel Prize for Physics in 1930.Raman

scattering produces scattered photons which differ in frequency from the radiation

source which causes it, and the difference is related to vibrational and/or rotational properties of the molecules from which the scattering occurs. It has become more

prominent in the years since powerful monochromatic laser sources could provide the

scattering power.

Raman scattering from lunar soil


Circular Orbit

Gravity supplies the necessary centripetal force to hold a satellite in orbit

about the earth. The circular orbit is a special case since orbits are generallyellipses, or hyperbolas in the case of objects which are merely deflected by

the planet's gravity but not captured. Setting the gravity force from the

univeral law of gravity equal to the required centripetal force yields the

description of the orbit. The orbit can be expressed in terms ofthe acceleration of gravity at the orbit.

Index

Gravity

concepts

Orbit

concepts









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The force of gravity in keeping an object in circular motion is an example

of centripetal force. Since it acts always perpendicular to the motion, gravitydoes not do work on the orbiting object if it is in a circular orbit.

Calculation

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Earth's Gravity

The weight of an object is given by W=mg, the force of gravity, which comes

from the law of gravity at the surface of the Earth in the inverse square

law form:

At standard sea level, the acceleration of gravity has the value g = 9.8 m/s2,

but that value diminishes according to the inverse square law at greater

distances from the earth. The value of g at any given height, say the height ofan orbit, can be calculated from the above expression.

Above the earth's surface at a height of h = m = x 106 m,

which corresponds to a radius r = x earth radius, the acceleration of gravity

is g = m/s2 = x g on the earth's surface.

Please note that the above calculation gives the correct value for the

acceleration of gravity only for positive values of h, i.e., for points outside the

Earth. If you drilled a hole through the center of the Earth, the acceleration ofgravity would decrease with the radius on the way to the center of the Earth.

If the Earth were of uniform density (which it is not!), the acceleration of

gravity would decrease linearly to half the surface value of g at half the radius

of the Earth and approach zero as you approached the center of the Earth.

Hole through center of Earth

Index

Gravity

concepts

Orbit

concepts


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Binary Circular Orbit

From the gravity force and the

necessary centripetal force:

If you are riding on one of the masses,

the relative motion equation has the

same form if you substitute

the reduced mass

which gives the orbit equation:

This leads to Kepler's 3rd law (the Law of Periods) which is useful for the

analysis of the orbits of moons and binary stars.

Since the period T of the orbit is given by

then the motion equation can be written

Index

Gravity

concepts

Orbitconcepts

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which reduces to

If we use the convenient astronomical units

r = a in AU (astronomical units)

G = 4π2

m in (solar masses)

T in years

then this takes the form

This applies to circular orbits where

a is the radius, and to elliptical orbits

where a is the semi-major axis.

and from just the period and orbit radius you can obtain the sum of the masses m1+

m2. If you can obtain the individual orbit radii r1 and r2 then you can use thecenter

of mass condition

with the measured mass sum to obtain the individual masses

These relationships are important in the study of visual binaries.


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Reduced Mass

The relative motion of two objects that are acted upon by a

central force can be described by Newton's 2nd Law as if

they were a single mass with a value called the "reduced

mass".

From Newton's 3rd Law :

The relative acceleration of the two masses is

Since a2 is negative, this can be rewritten in terms of the magnitudes of the quantities:


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Force from spherical shell of mass M

Gravitational lens

Inde

x

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Gravity

Gravity is the weakest of the four fundamental forces, yet it is the dominant force in the

universe for shaping the large scale structure of galaxies, stars, etc. The gravitational

force between two masses m1 and m2 is given by the relationship:

This is often called the "universal law of gravitation" and G the universal gravitation

constant. It is an example of an inverse square law force. The force is always attractiveand acts along the line joining the centers of mass of the two masses. The forces on the

two masses are equal in size but opposite in direction, obeying Newton's third law.

Viewed as an exchange force, the massless exchange particle is called the graviton.

The gravity force has the same form as Coulomb's law for the forces between electric

charges, i.e., it is an inverse square law force which depends upon the product of thetwo interacting sources. This led Einstein to start with the electromagnetic force and

gravity as the first attempt to demonstrate the unification of the fundamental forces. It

turns out that this was the wrong place to start, and that gravity will be the last of theforces to unify with the other three forces. Electroweak unification (unification of the

electromagnetic and weak forces) was demonstrated in 1983, a result which could not

be anticipated in the time of Einstein's search. It now appears that the common form of

the gravity and electromagnetic forces arises from the fact that each of them involves

an exchange particle of zero mass, not because of an inherent symmetry which wouldmake them easy to unify.

Index

Gravity

Concept

s

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Examples of Trajectories

Common misconceptions about guns:

A dropped bullet will hit the ground before one which is fired from a

gun.

As shown in the illustration of a horizontal launch, gravity acts the same way on

both bullets, giving them the same downward acceleration and making them strike

the ground at the same time if the bullet is fired horizontally over level ground.

Bullets fired from high-powered rifles drop only a few inches in

hundreds of yards.

Fired at twice the speed of sound, a bullet will drop over 3 inches in 100 yards, and

at 300 yards downrange will have dropped about 30 inches. Plug in numbers into

the bullet drop calculation to see for yourself. Ammunition manufacturers

contribute to this misconception by stating the drop of their projectiles as just the

extra drop caused by frictional drag compared to an ideal frictionless projectile.

Index

Trajectory

concepts


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Drop of a Bullet

If air friction is neglected, then the drop of a bullet fired horizontally can betreated as an ordinary horizontal trajectory. The air friction is significant, so

this is an idealization.

If the muzzle velocity is

v = m/s = ft/s = mi/hr = km/hr

and the distance downrange is

R = m = ft = yards,

Then the amount of drop of the bullet below the horizontal would be

d = m = cm = inches

If the gun is fired on level ground at a height of m = ft, then

the bullet will hit the ground in seconds, having traveled a distance

of meters = feet.

To hold the drop to cm = inches at the downrange distance R

above would require a muzzle velocity of m/s = ft/s.

Index

Trajectoryconcepts















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Index

Motion

concept

s




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Freefall

In the absence of frictional drag, an object near the surfaceof the earth will fall with the constant acceleration of

gravity g. Position and speed at any time can be calculatedfrom themotion equations.

Illustrated here is the situation where an object is released

from rest. It's position and speed can be predicted for anytime after that. Since all the quantities are directeddownward, that direction is chosen as the positive

direction in this case.

At time t = s after being dropped,

the speed is vy = m/s = ft/s ,

The distance from the starting point will be

y = m= ft

Enter data in any box and click outside the box.

Note that you can enter a distance (height) and click outside the box to calculate the freefall

time and impact velocity in the absence of air friction. But the calculation assumes that the

gravity acceleration is the surface value g = 9.8 m/s2, so the height is great enough for gravity

to have changed significantly the results will be incorrect.

Free fall with air friction

Free fall from great height

Free fall experiment with spark timer

Index

Trajectory

concepts


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Peak at m at

t=

s

Vertical Trajectory

Vertical motion under the influence of gravity can be described by the

basic motion equations. Given the constant acceleration of gravity g,

the position and speed at any time can be calculated from the motion

equations:

You may enter values for launch velocity and time in the boxes below

and click outside the box to perform the calculation.

For launch speed v0y = m/s = ft/s

and time t = s ,

The values below are output values; those boxes will not accept input for

calculation. The velocity will be

vy = m/s = ft/s

and the height will be y = m = ft

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Horizontal Launch

All the parameters of a horizontal launch can be calculated with the motion equations,

assuming a downward acceleration of gravity of 9.8 m/s2.

Time of flight

t = s

Vertical impact velocity

vy = m/s

Launch velocity

v0 = m/s

Height of launch

h = m

Horizontal range

R = m

Calculation is initiated by

clicking on the formula inthe illustration for thequantity you wish to

calculate.

Include demonstration apparatus

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General Ballistic Trajectory

The motion of an object under the influence of gravity is determined completely by the acceleration

of gravity, its launch speed, and launch angle provided air friction is negligible. The horizontal and

vertical motions may be separated and described by the general motion equations for constantacceleration. The initial vector components of the velocity are used in the equations. The diagram

shows trajectories with the same launch speed but different launch angles. Note that the 60 and 30

degree trajectories have the same range, as do any pair of launches at complementary angles. The

launch at 45 degrees gives the maximum range.

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For launch velocity v0 = m/s, launch angle θ

= degrees:

At time t = sec:

Horizontal velocity

vx = m/s.

Horizontal distance

x = m.

Vertical velocity

vy = m/s.

Vertical position

y = m.

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For launch velocity

v0 = m/s,

launch angle

θ = degrees,

The horizontal range is

R = m.

The total time of flight is

t = s.

The peak height is

h = m.

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Will it clear the fence?

The basic motion equations can be solved simultaneously to express y in terms of x.

For launch velocity

v0 = m/s =

ft/s, launch angle

θ = degrees,

and horizontal range

x = m = ft,

the calculated height is

y = m = ft.

The time of flight is

t = s.

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Where will it land?

The basic motion equations give the position components x and y in terms of the time. Solving for

the horizontal distance in terms of the height y is useful for calculating ranges in situations where

the launch point is not at the same level as the landing point.

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Where will it land?

The basic motion equations give the position components x and y in terms of the time.

Solving for the horizontal distance in terms of the height y is useful for calculating ranges in

situations where the launch point is not at the same level as the landing point.

Launch velocity

v0 = m/s = ft/s,

launch angle

θ = degrees,

and trajectory height

y = m = ft,

The two calculated times are

t1 = s and

t2 = s.

The corresponding ranges are

x1 = m = ft

and

x2 = m = ft.

Note that the value y in the illustration is downward

and it is presumed that upward is positive. To

reproduce the scenario in the diagram, the input value

of y should be negative.

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Launch Velocity

The launch velocity of a projectile can be calculated from the range if the angle of launch is

known. It can also be calculated if the maximum height and range are known, because theangle can be determined.

From the range relationship,

the launch velocity can be

calculated. For range

R = m = ft,

and launch angle

θ = degrees,

the launch velocity is

v0 = m/s =

ft/s.

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Launch Velocity

The launch velocity of a projectile can be calculated from the range if the angle of launch is

known. It can also be calculated if the maximum height and range are known, because the

angle can be determined.

For range

R = m = ft,

and peak height

h = m = ft,

the launch velocity is

v0 m/s = ft/s.

The required launch angle is

θ = degrees.

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Angle of Launch

Variation of the launch angle of a projectile will change the range. If the launch velocity is

known, the required angle of launch for a desired range can be calculated from the motionequations.

From the range relationship,

the angle of launch can be

determined. For range

R = m = ft,

and launch velocity

v0 = m/s =

ft/s.

there are two solutions for the

launch angle.

θ1 = degrees,

θ2 = degrees,

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