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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.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3
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!
http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html#c2
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http://hyperphysics.phy-astr.gsu.edu/hbase/grexp.html#c1
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.
http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c1
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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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.
http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibspe.html#c1
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
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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.
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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
Molecule Dissociation Energy(eV)Equilibrium Separation (nm)
(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
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diatomic.html#c2
Homonuclear Diatomic Molecules
Molecule Dissociation Energy(eV)Equilibrium Separation (nm)
(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
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diatomic.html#c2
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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
http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/raman.html#c1
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:
http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#rm
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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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Note: This is a large HTML document. Wait until it finishes loading for full functionality.
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
Index
Trajectory
concepts
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Index
Trajectory
concepts
Calculatio
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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
Index
Trajectory
concepts
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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.
Index
Trajectory
concepts
Calculatio
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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.
Index
Trajectory
concepts
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Index
Trajectory
concepts
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Index
Trajectory
concepts
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Index
Trajectory
concepts
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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.
Index
Trajectory
concepts
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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.
Index
Trajectory
concepts
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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.
Index
Trajectory
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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.
Index
Trajectory
concepts
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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.
Index
Trajectory
concepts
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8/11/2019 Random Physics articles
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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.
Index
Trajectory
concepts
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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,
Index
Trajectory
concepts
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8/11/2019 Random Physics articles
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http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tracon