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| faizankhan666@gmail.com |
 Posted: Thu Sep 14, 2006 10:50 am |
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Guest
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A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Please, Kindly give me all the formulas used in the solution of this
question. |
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| Sam Wormley |
| Posted: Thu Sep 14, 2006 10:57 am |
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faizankhan666@gmail.com wrote:
[quote:88588caa19]A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Please, Kindly give me all the formulas used in the solution of this
question.
[/quote:88588caa19]
How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive. |
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| faizankhan666@gmail.com |
| Posted: Thu Sep 14, 2006 11:27 am |
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Guest
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[quote:b6f46510c7]How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive.
[/quote:b6f46510c7]
No give me full solution or give me formulas. |
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| Sam Wormley |
| Posted: Thu Sep 14, 2006 11:46 am |
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Guest
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faizankhan666@gmail.com wrote:
[quote:1c2f6fbbd3]
How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive.
No give me full solution or give me formulas.
[/quote:1c2f6fbbd3]
Start with Newton's
Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Let's use the case where force is constant:
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o |
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| Helmut Wabnig |
| Posted: Thu Sep 14, 2006 12:03 pm |
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On Thu, 14 Sep 2006 13:57:05 GMT, Sam Wormley <swormley1@mchsi.com>
wrote:
[quote:a6b174d5be]faizankhan666@gmail.com wrote:
A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Please, Kindly give me all the formulas used in the solution of this
question.
How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive.
he said "high..rifle...be pointed" not elevated, assume same height,[/quote:a6b174d5be]
the usual parabola calculation, without air resistance, wind,
and temperature inversions.
http://www.du.edu/~jcalvert/math/parabola.htm
http://en.wikipedia.org/wiki/Trajectory
http://www.usdidactic.com/refpdf/scipdf/Experim/1_3_11.pdf#search=%22projectile%20inclination%20formula%20parabola%22
w. |
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| Guest |
| Posted: Thu Sep 14, 2006 12:35 pm |
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Sam Wormley <swormley1@mchsi.com> wrote:
[quote:0d614ee899]faizankhan666@gmail.com wrote:
How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive.
No give me full solution or give me formulas.
Start with Newton's
Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Let's use the case where force is constant:
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
[/quote:0d614ee899]
Now you need to work in the ballistic coefficient which is in essense
a drag number for bullets that takes into account the overall shape,
size, and length to diameter ratio.
Oh, and don't forget to account for air pressure, temperature and
humidity.
Or go get a reloading manual and look up the bullet in the ballistics
tables.
--
Jim Pennino
Remove .spam.sux to reply. |
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| Greg Neill |
| Posted: Thu Sep 14, 2006 3:21 pm |
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Guest
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<faizankhan666@gmail.com> wrote in message
news:1158241839.504869.31430@k70g2000cwa.googlegroups.com...
[quote:857d489d2c]
A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Please, Kindly give me all the formulas used in the solution of this
question.
[/quote:857d489d2c]
You might get some help for your homework if you
would at least show that you've put in some effort
yourself. Show us what you've tried and where
you're stuck. |
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| Timo Nieminen |
| Posted: Thu Sep 14, 2006 7:28 pm |
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Guest
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On Thu, 14 Sep 2006, faizankhan666@gmail.com wrote:
[quote:acd0d31a60]A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
[/quote:acd0d31a60]
Coincidentally, I was telling a student about the practical application of
this problem yesterday. This was with a rather feeble air rifle with a
muzzle velocity of 44 m/s (measured by shooting straight up an waiting for
it to come back down). Target was about 50 metres away, a 200l drum (aka
44 gallon drum). Knowing the muzzle velocity, I knew it would take about
1s for the pellet to reach, and knowing y= gt^2/2, obviously one needs to
aim about 5m high. Did so. Shoot, and one second later, "ting!"
[quote:acd0d31a60]Please, Kindly give me all the formulas used in the solution of this
question.
[/quote:acd0d31a60]
These you should already have:
"equation of motion in 1D for constant velocity"
"equation of motion in 1D for constant acceleration".
That's all you need.
You should also consider the point of being required to answer questions
like the above. Traditionally, introductory physics courses are concluded
by an exam wherein the student needs to solve a certain number of problems
such as the above in a limited time, with limited resources. If you
actually do some problems like them beforehand, you'll find it much easier
to do them in the exam. Reading somebody else's answer and writing it down
just doesn't give you the same practice. If the above question is
compulsory homework, it's compulsory because the teachers found in the
past that if it wasn't compulsory, too many students didn't do it for
homework, and as a consequence, couldn't do it in the exam.
If you're being forced to take a physics course you don't want to take,
and you want to pass it with minimum effort, don't bother taking shortcuts
when given homework. Do it properly, and you'll spend less time cramming
for the exam. Put the time in now, when you can ask your
lecturer/teacher/tutor about the details, rather than in the 3 days before
the exam when you might not be able to ask them. You should get the past
exams for the last 5 years, and do the questions on them as you cover the
material in class, and ask about anything you have problems with as you
go.
If you're doing the course to actually learn some physics, then do so.
Asking to be spoonfed won't do that. Do and learn. It is possible to pass
introductory physics exams without understanding any physics, simply by
memorising solutions to the most likely exam questions. IMHO, such
memorising actually takes more effort than learning the tiny handful of
principles of physics underlying those questions, and the basics of
problem solving techniques. It's your choice, and you might find the
former (ie memorisation) easer. But if you actually want to learn physics,
do it properly.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| tadchem |
| Posted: Thu Sep 14, 2006 9:35 pm |
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faizankhan666@gmail.com wrote:
[quote:45254dc6f7]How long does it take the bullet to get to the target?
How far does the bullet fall in that time?
Either use formulae from your textbook, or derive them yourself,
which would be far more instructive.
No give me full solution or give me formulas.
[/quote:45254dc6f7]
OK.
http://en.wikipedia.org/wiki/List_of_equations_in_classical_mechanics
There is probably something there you can use, ya spoiled brat.
Are you smart enough to figure out *which* equations you need and which
you don't need?
Let us know your answer - I've got money riding on this...
Tom Davidson
Richmond, VA |
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| Greg Neill |
| Posted: Thu Sep 14, 2006 10:41 pm |
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"Timo Nieminen" <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0609150808200.32764-100000@localhost...
[quote:1dd3229927]On Thu, 14 Sep 2006, faizankhan666@gmail.com wrote:
A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Coincidentally, I was telling a student about the practical application of
this problem yesterday. This was with a rather feeble air rifle with a
muzzle velocity of 44 m/s (measured by shooting straight up an waiting for
it to come back down). Target was about 50 metres away, a 200l drum (aka
44 gallon drum). Knowing the muzzle velocity, I knew it would take about
1s for the pellet to reach, and knowing y= gt^2/2, obviously one needs to
aim about 5m high. Did so. Shoot, and one second later, "ting!"
[/quote:1dd3229927]
Keeping in mind, of course, that in general there are *two*
angles that will satisfy the conditions... |
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| Ben newsam |
| Posted: Fri Sep 15, 2006 6:03 am |
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On Fri, 15 Sep 2006 08:28:20 +1000, Timo Nieminen
<timo@physics.uq.edu.au> wrote:
[quote:82de628575]Coincidentally, I was telling a student about the practical application of
this problem yesterday. This was with a rather feeble air rifle with a
muzzle velocity of 44 m/s (measured by shooting straight up an waiting for
it to come back down). Target was about 50 metres away, a 200l drum (aka
44 gallon drum). Knowing the muzzle velocity, I knew it would take about
1s for the pellet to reach, and knowing y= gt^2/2, obviously one needs to
aim about 5m high. Did so. Shoot, and one second later, "ting!"
[/quote:82de628575]
With a muzzle velocity of 44 m/s, I think you were doing quite well to
hit something 50 m away after 1 second.
--
Posted via a free Usenet account from http://www.teranews.com |
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| Timo A. Nieminen |
| Posted: Sun Sep 17, 2006 5:13 pm |
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On Thu, 14 Sep 2006, Greg Neill wrote:
[quote:52a5d7a514]"Timo Nieminen" <timo@physics.uq.edu.au> wrote:
On Thu, 14 Sep 2006, faizankhan666@gmail.com wrote:
A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7
m away and level with the rifle.How high above the target must the
rifle barrel be pointed so that the bullet hits the target?
Coincidentally, I was telling a student about the practical application of
this problem yesterday. This was with a rather feeble air rifle with a
muzzle velocity of 44 m/s (measured by shooting straight up an waiting for
it to come back down). Target was about 50 metres away, a 200l drum (aka
44 gallon drum). Knowing the muzzle velocity, I knew it would take about
1s for the pellet to reach, and knowing y= gt^2/2, obviously one needs to
aim about 5m high. Did so. Shoot, and one second later, "ting!"
Keeping in mind, of course, that in general there are *two*
angles that will satisfy the conditions...
[/quote:52a5d7a514]
Indeed, but in practice, one angle is easier than the other. Sometimes
even the high angle, given obstructions.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Timo A. Nieminen |
| Posted: Sun Sep 17, 2006 5:15 pm |
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Guest
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On Fri, 15 Sep 2006, Ben newsam wrote:
[quote:efc897b897]Timo Nieminen <timo@physics.uq.edu.au> wrote:
Coincidentally, I was telling a student about the practical application of
this problem yesterday. This was with a rather feeble air rifle with a
muzzle velocity of 44 m/s (measured by shooting straight up an waiting for
it to come back down). Target was about 50 metres away, a 200l drum (aka
44 gallon drum). Knowing the muzzle velocity, I knew it would take about
1s for the pellet to reach, and knowing y= gt^2/2, obviously one needs to
aim about 5m high. Did so. Shoot, and one second later, "ting!"
With a muzzle velocity of 44 m/s, I think you were doing quite well to
hit something 50 m away after 1 second.
[/quote:efc897b897]
The Benefits of Physics! And 200l drums aren't small.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| tadchem |
| Posted: Sat Sep 30, 2006 7:14 am |
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Timo A. Nieminen wrote:
[quote:d623cdecf0]On Thu, 14 Sep 2006, Greg Neill wrote:
Keeping in mind, of course, that in general there are *two*
angles that will satisfy the conditions...
Indeed, but in practice, one angle is easier than the other. Sometimes
even the high angle, given obstructions.
[/quote:d623cdecf0]
The choice is often made by comparing the effects of different angles
of impact.
A lobbed projectile (taking the high angle - called 'indirect fire' by
artillerists) is often more effective against targets that are weakly
armored against aerial attacks.
Direct fire is often more accurate as it spends less time in the air
and stays at lower altitudes, making it less subject to 'windage'.
Tom Davidson
Richmond, VA |
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