Projectile motion with friction?

In summary: The OP was asked to decide on the form of the drag and responded with the linearised form. Therefore, my response was based on the OP's decision to consider linear drag.
  • #1
loup
36
0
If there is air resistance, when projectile motion occurs, in what angle we throw the object can we obtain the highest range?

Can anybody please help me with this question? I have been doubtful about this question. My teacher says it has something to do with calculus but he has already forgotten some rules of calculas.

Can anybody answer this question? If you show how you can arrive this question, it will be better. Thank you.
 
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  • #2
It depends on the form of the air resistance contribution to the force.
 
  • #3
...but, it the resistant force is assumed linear in the velocity the I think the problem isn't too hard to work out. Write down the equations of motion and we can work through it together if you want. Cheers.
 
  • #4
I think the answer is that nobody knows, but folks around here seem to disagree with me. I think that the flight of an object through a gas is highly non-trivial. The two prevailing approximations are that the force of resistance due to drag from the gas is proportional to either the first power or the square of the velocity, with the proportionality constant determined from experiment. To truly do this problem I think you would need to solve a compressible Navier-Stokes setup. CrazY!
 
  • #5
About 34 degrees
 
  • #6
Hey, How Phrak works the answer out?

Olgranpappy, Here is the equation:
mg+bv=ma for veritcal motion of the thing.

bv=ma for horizontal motion.

I sincerely believe that the result requires something to do with calculus, can anybody who is excellent in calculus help me to deal with the problem?
 
  • #7
loup said:
Hey, How Phrak works the answer out?

Olgranpappy, Here is the equation:
mg+bv=ma for veritcal motion of the thing.

bv=ma for horizontal motion.

I sincerely believe that the result requires something to do with calculus, can anybody who is excellent in calculus help me to deal with the problem?
Okay, let's start with the vertical component. Acceleration is the second derivative of position with respect to time and velocity if the first derivative of position with respect to time. Hence, your equation may be written

[tex]m\frac{d^2y}{dt^2} = b\frac{dy}{dt} + mg[/tex]

or in canonical form:

[tex]\frac{d^2y}{dt^2} - \frac{b}{m}\frac{dy}{dt} = g[/tex]

Which is a second order linear ODE with constant coefficients. Have you met such equations before?

As this question is certainly in the style of a homework question, I'm moving it to Into. Phys. in the Homework forums.
 
  • #8
loup said:
Hey, How Phrak works the answer out?
I didn't work it out. I was being flippant. Including air friction, the angle will always be somewhere less than 45 degrees in still air. How many degrees is dependent upon a lot of conditions including Reynolds number.

@Hootenanny. Viscous drag occurs at very low speeds and is proportional to the velocity. But in most worldly cases you'd be dealing with turbulent drag, proportional to the square of the velocity, as olgranpappy may have been implying.

[tex] m \ddot {x} = -mg - c \dot{x}^2 [/tex]

https://www.physicsforums.com/newreply.php?do=newreply&p=816101" [Broken]
 
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  • #9
Phrak said:
@Hootenanny. Viscous drag occurs at very low speeds and is proportional to the velocity. But in most worldly cases you'd be dealing with turbulent drag, proportional to the square of the velocity, as olgranpappy was implying.

[tex] m \ddot {x} = -mg - c \dot{x}^2 [/tex]

https://www.physicsforums.com/newreply.php?do=newreply&p=816101" [Broken]
Yes I know. However, it is entirely up to the OP how they want to define the parameters of the problem. The OP has already been made aware that the result is highly dependent upon the form of the drag, whether quadratic or otherwise.

The OP was asked to decide on the form of the drag and responded with the linearised form. Therefore, my response was based on the OP's decision to consider linear drag.
 
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  • #10
loup said:
I sincerely believe that the result requires something to do with calculus, can anybody who is excellent in calculus help me to deal with the problem?

It certainly does have something to do with calculus--namely, we have to solve differential equations.

In order to make our problem well defined I think that the assumption of a linear drag force is fine. In this case we will be able to solve the equations of motion exactly. Unfortunately, we will find that we cannot solve for the range (at least I couldn't) in terms of simple well-known functions.
 
  • #11
Hootenanny said:
Okay, let's start with the vertical component. Acceleration is the second derivative of position with respect to time and velocity if the first derivative of position with respect to time. Hence, your equation may be written

[tex]m\frac{d^2y}{dt^2} = b\frac{dy}{dt} + mg[/tex]

There may be a sign error in the above equation. I think, +mg should be -mg... assuming that we take "up" as positive... isn't this usually what we do? I guess it's a convention... Also, I usually prefer to define constants such as "b" without any hidden minus signs, so I would probably flip the sign on "b" as well.

But regardless it will be good to know whether the OP has seen linear differential equations before.
 
  • #12
olgranpappy said:
There may be a sign error in the above equation. I think, +mg should be -mg... assuming that we take "up" as positive... isn't this usually what we do? I guess it's a convention... Also, I usually prefer to define constants such as "b" without any hidden minus signs, so I would probably flip the sign on "b" as well.

But regardless it will be good to know whether the OP has seen linear differential equations before.
It's purely a matter of taste for me. I usually choose all my coefficients such that all the signs in the DE are positive, but I concede that it can be confusing, particularly when trying to keep a track of directions. So yes, it would probably be more informative to re-write the ODE

[tex]\ddot{y}-b\dot{y} = -mg[/tex]

Where b,m,g are positive constants.
 
  • #13
ODE....
I have not learned it yet.

Are you two sure that it is less than 45 degree for the max. range? Why you cannot work out the range?What is the final equation you obtain? Is it involved both cos and sin?
 
  • #14
loup said:
ODE....
I have not learned it yet.

Are you two sure that it is less than 45 degree for the max. range? Why you cannot work out the range?

The range can be worked out, but doing that requires solving an Ordinary Differential Equation (ODE), which you haven't learned about yet.

What is the final equation you obtain? Is it involved both cos and sin?

The idea of this forum is to help YOU obtain the equation, not to have other people do all the work for you.

But honestly, I am having trouble seeing the benefit in assigning a differential equations problem to students who are just beginning calculus. This is more a comment about your teacher, and not about you personally.
 
  • #15
Redbelly98 said:
But honestly, I am having trouble seeing the benefit in assigning a differential equations problem to students who are just beginning calculus, or worse have not had any calculus. This is more a comment about your teacher, and not about you personally.
I totally agree. In fact, I was shocked to read that
loup said:
My teacher says it has something to do with calculus but he has already forgotten some rules of calculas.
This is indeed worrying. I'm appalled that someone teaching what appears to be college level physics has forgotten calculus and doesn't appear to know how to solve basic ODE's beyond that it has something to do with calculus.
 
  • #16
Loup, this isn't a homework assignment, is it?
What class is this for, and what grade are you in?
 
  • #17
loup said:
ODE....
I have not learned it yet.

Are you two sure that it is less than 45 degree for the max. range? Why you cannot work out the range?What is the final equation you obtain? Is it involved both cos and sin?

well... since this isn't really a homework problem maybe it is okay to give a few equations explictly as examples instead of being so very Socratic...

For example, to find the time of flight (T) you would have to solve the equation:
[tex]
0=v_0\sin(\phi)\left(\frac{1-e^{(-\beta T)}}{\beta}\right)-\frac{g}{\beta}\left(T-\frac{1-e^{(-\beta T)}}{\beta}\right)\;,
[/tex]
where [itex]\phi[/itex] is the angle that the launcher makes with the horizontal, and where g is 9.8m/s^2 and where [itex]\beta=\alpha/m[/itex], where [itex]\alpha[/itex] determines the resistant force as
[tex]
\bold{F}_{\rm res}=-\alpha\bold{v}\;,
[/tex]
where v is the velocity and v_0 is the initial velocity.

The problem is not just the that equation is messy, the problem is that the equation involves T in both a polynomial and an exponential and so cannot be solved for in terms of elementary functions. But, it is possible to do it numerically without too much effort.

Then, given the time of flight one can plug into a similar equation for the x-position to find the range. And then take a dirivative with respect to \phi and set that equal to zero to find the angle which maximizes the range.

Cheers.
 
  • #18
Msg me if you want the equation for time to reach termial velocity or distance. Just did it a hour ago, took me 4 days...
 
  • #19
olgranpappy, you are kind. Indeed it is not a homework question. It is simply for my interest. The exams or tests will even not ask this kind of questions. It is just too difficult. I am a Secondary 6 student in Hong Kong. I don't know how to translate that to grade to make you understand :(

To others, Don't you blame my teacher. He is a nice guy, Since it is not in the syllabus, the teacher hasn't done this kind of Mathematics for ages. It is normal to forget. It is not he forgot how to do basic calculus, but the more deep calculus like ODE.

In fact, I know about the 2 equations but I don;t know how to solve them. My teacher says it is less than 45 degrees and I doubt. People shouldn't believe something unless proof is given. That's why I ask the question here, in order to find someone who can solve the equations for me. If it is something I can cope with, I will have no need to ask the question here. By the way, I did raise the 2 equations myself!

Here in Hong Kong, peole who learn Physics have no need to learn about Pure Mathematics. I take the course, Physics, Chemistry and Biology. I am reading a Pure Mathematics books myself in order to cope with the difficutlies I met in Physics. But it requries time and before I am finish the books, I need you guys to help me. Thanks a lot for your help.

Maybe you will think I am rather rude in this message, am I ? Please forgive me, I am not good at speaking and talking to others.
 
  • #20
If it's not an assigned homework problem, then that is another matter. No, you're not being rude, and it sounds like you have a good teacher.

I've been able to verify olgranpappy's method, so yes the solution requires numerical methods. There is no equation using familiar functions that will provide the answer.

Bright Wang said:
Msg me if you want the equation for time to reach termial velocity or distance. Just did it a hour ago, took me 4 days...

Since terminal velocity is never reached, can you clarify what you really mean?
 
  • #21
Redbelly98 said:
If it's not an assigned homework problem, then that is another matter. No, you're not being rude, and it sounds like you have a good teacher.

I've been able to verify olgranpappy's method, so yes the solution requires numerical methods. There is no equation using familiar functions that will provide the answer.



Since terminal velocity is never reached, can you clarify what you really mean?

lol I mean approxmatly. This there a way to post pic on this?? I have the LaTex in pic form.
 
  • #22
loup said:
olgranpappy, you are kind. Indeed it is not a homework question. It is simply for my interest. The exams or tests will even not ask this kind of questions. It is just too difficult. I am a Secondary 6 student in Hong Kong. I don't know how to translate that to grade to make you understand :(

To others, Don't you blame my teacher. He is a nice guy, Since it is not in the syllabus, the teacher hasn't done this kind of Mathematics for ages. It is normal to forget. It is not he forgot how to do basic calculus, but the more deep calculus like ODE.

In fact, I know about the 2 equations but I don;t know how to solve them. My teacher says it is less than 45 degrees and I doubt. People shouldn't believe something unless proof is given.

A simple proof that doesn't rely on solving equations would be to go outside and try throwing a baseball (or better yet a softball) at an angle of about 30 degrees, about 45 degrees, and about 60 degrees and then check which throw went the farthest.

Actually, in this case, the air resistance is more well approximated as proportional to the square of the velocity rather than linear in the velocity.

Here is a web page that does some numerical calculation for the case when the air resistance is proportional to the square of the velocity:
http://physics.ucsc.edu/~peter/115/range.nb.pdf [Broken]

That's why I ask the question here, in order to find someone who can solve the equations for me. If it is something I can cope with, I will have no need to ask the question here. By the way, I did raise the 2 equations myself!

Here in Hong Kong, peole who learn Physics have no need to learn about Pure Mathematics. I take the course, Physics, Chemistry and Biology. I am reading a Pure Mathematics books myself in order to cope with the difficutlies I met in Physics. But it requries time and before I am finish the books, I need you guys to help me. Thanks a lot for your help.

Maybe you will think I am rather rude in this message, am I ? Please forgive me, I am not good at speaking and talking to others.
 
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  • #23
Googling "projectile motion simulator" delivered plenty of hits. Here's one:

http://www.walter-fendt.de/ph11e/projectile.htm" [Broken]

On second glance, this one ignores air friction, but there are others...
 
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  • #24
Bright Wang said:
lol I mean approxmatly. This there a way to post pic on this?? I have the LaTex in pic form.

You can put the image up on an image-hosting service (for example http://www.photobucket.com ), then display it here by typing:

[noparse][/noparse][B][COLOR="Blue"]URL of image goes here[/COLOR][/B] [Broken]

(That should all be typed without pressing the Enter key)
 
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  • #25
[tex] t=\frac{1}{k\sqrt{g/K}}\ \ln[\ \frac{\sqrt{g/K} + \sqrt{g/K- \frac{(g-Kv_{o}^2)\times e^-^2^K^y}{K}}} {\sqrt{\frac{(g-Kv_{o}^2)}{K}}\times e^-^k^y} \ ] [/tex]

here's the relationship between time and y(distance). where [tex] K=\frac {pC_{d}A}{2m} [/tex]

[tex] y=\frac{-1}{2K} \times \ln [ \frac {g-Kv_{f}^2}{g-Kv_{o}^2} ] [/tex]

and here's the relationship between distance and vo and vf.

Note: only the y-axis is considered. If not the equation would be interwined, not sure how to solve that? (complex numbers?) anyways I'm only in high school...
 
  • #26
Actually, I really thank you all for really thinking about my question. I really enjoy having this forum. Thanks for the contributor of this forum. I always consider discussing Physics problem interesting and fun.
 
  • #27
Bright Wang said:
[tex] t=\frac{1}{k\sqrt{g/K}}\ \ln[\ \frac{\sqrt{g/K} + \sqrt{g/K- \frac{(g-Kv_{o}^2)\times e^-^2^K^y}{K}}} {\sqrt{\frac{(g-Kv_{o}^2)}{K}}\times e^-^k^y} \ ] [/tex]

here's the relationship between time and y(distance). where [tex] K=\frac {pC_{d}A}{2m} [/tex]

[tex] y=\frac{-1}{2K} \times \ln [ \frac {g-Kv_{f}^2}{g-Kv_{o}^2} ] [/tex]

and here's the relationship between distance and vo and vf.

Note: only the y-axis is considered. If not the equation would be interwined, not sure how to solve that? (complex numbers?) anyways I'm only in high school...

In trying to understand your equations, I have several questions:

What are p, Cd, and A?
Is k the same as K, or is it something else?
Is vf the terminal velocity, or the velocity when the projectile hits the ground, or something else?
Where is the dependence on launch angle?
 
  • #28
Redbelly98 said:
In trying to understand your equations, I have several questions:

What are p, Cd, and A?
Is k the same as K, or is it something else?
Is vf the terminal velocity, or the velocity when the projectile hits the ground, or something else?
Where is the dependence on launch angle?

p is the density of the medium. Cd is the drag coefficient, and A is the cross sectional area. I'm not doing this problem, this is equations for a falling object on the y-axis. Vf would be the velocity when it hits the ground, that also can be the termial velocity. This equation can solve how long it takes for a object to reach termial velocity (like if you just out of a air plane) or the distance. I don't know how to do this problem with a angle, there would be two equations but they would be interwined (or some other word). and then still use the ODE?? Any ways I'm just trying to say this is not a easy problem.

There was someone who said 34 degree or something. how did he get that?
 
  • #29
Okay, thanks for clarifying.

Bright Wang said:
There was someone who said 34 degree or something. how did he get that?

He was joking. We only know it's less than 45 degrees if there is any drag.
 
  • #30
Bright Wang said:
p is the density of the medium. Cd is the drag coefficient, and A is the cross sectional area. I'm not doing this problem,

Then don't post in the thread. It's distracting.

this is equations for a falling object on the y-axis. Vf would be the velocity when it hits the ground, that also can be the termial velocity. This equation can solve how long it takes for a object to reach termial velocity

You don't need an equation for that. Terminal velocity is not reached at any finite time.

(like if you just out of a air plane) or the distance. I don't know how to do this problem with a angle, there would be two equations but they would be interwined (or some other word). and then still use the ODE?? Any ways I'm just trying to say this is not a easy problem.

There was someone who said 34 degree or something. how did he get that?

He didn't, he was just trying to be funny. See Post #8.
 
  • #31
Loup.

Here is a fairly good animation you can run.

"[URL [Broken]
http://www.compadre.org/PSRC/items/detail.cfm?ID=7196

Click on Modular Approach to Physics: Projectile Motion: One Ball.

It will give you a good intuitive understanding of the problem.

I recommend making the grid, trace, and velocity visible with the buttons. Drag the end of the velocity arrow around keeping it at 100 m/s. Change the drag coefficient to 0.01[corrected] or the air drag will not be significant. In this applet the drag is proportional to v squared, as it should be, and if their language used in the applet is correct.
 
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  • #32
!
I cannot open any of the animation!
:(
 
  • #33
Phrak, good find! But I think you meant 0.01 for the drag coef.

You can also enter numbers for the velocity and angle, but you must hit the "Enter" key for them to register.
 
  • #34
loup said:
!
I cannot open any of the animation!
:(

You may need to install java, if your computer does not already have it.
http://www.java.com/en/download/manual.jsp [Broken]
 
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  • #35
Redbelly98 said:
Phrak, good find! But I think you meant 0.01 for the drag coef.

You can also enter numbers for the velocity and angle, but you must hit the "Enter" key for them to register.

Thanks on all points, Redbelly. I was much to critical of the software--it's excellent. The use of the energy key makes all the difference.
 

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