Is there a quicker way to solve this projectile problem?

  • Thread starter Jahnavi
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Homework Statement


viscous.jpg


Homework Equations




The Attempt at a Solution



I solved this problem and got the correct answer 108 m . But I had to solve a DE . This question is from a test where we get around 1 minute to solve a problem . I took half an hour to solve it :wideeyed: .

It looks like a subjective type problem .

I am just curious as to whether there is a quicker/smarter way to find one of the options .
 

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  • #2
Charles Link
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What if you considered just the horizontal motion and the viscous force acting on the horizontal component of the velocity? That way, it still needs a differential equation, but only in one dimension. ## \\ ## It's a very simple D.E. that takes at most 2 minutes to solve for ## v(t) ## and then you can integrate ## v(t) ## from ## t=0 ## to ## t=10 ##.## \\ ## Editing: Yes, I get ## s=108 ## m also. ## \\ ## And note: They make it easy by making ## m=1 \, kg ##, so that ## a=\frac{dv}{dt}=\frac{F}{m}=\frac{-.2v}{1}=-.2 v ##. ## \\ ## Question: Is this the D.E. that you used, or did you try to include the vertical motion?
 
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  • #3
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What if you considered just the horizontal motion and the viscous force acting on the horizontal component of the velocity? That way, it still needs a differential equation, but only in one dimension.
I did the same :smile:

Is this the D.E. that you used,
Yes !
 
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  • #4
Charles Link
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I did the same :smile:



Yes !
The first time you see that D.E. it might take 15 minutes or more to solve, but that is one that I think I have seen hundreds of times, and it is a simple matter to write: ## \frac{dv}{v}=-\lambda \, dt ## ==>> ## \ln| v|=-\lambda t +C ## so that ## v(t)=v_o e^{- \lambda t} ##. The first several times it probably took me a few minutes or longer...
 
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It's a very simple D.E. that takes at most 2 minutes
The first time you see that D.E. it might take 15 minutes or more to solve
I can see , you are trying to make me feel better :biggrin: .

Nevertheless , my intent of posting this question was to know if some of the options could be eliminated and correct answer could be chosen without employing calculus .
 
  • #6
Charles Link
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Nevertheless , my intent of posting this question was to know if some of the options could be eliminated and correct answer could be chosen without employing calculus .
I don't think there is any way besides calculus, but the calculus solution, including the solution of the D.E., and integrating ## v(t) ## is very quick=quick enough that if you spent a minute or two looking for something simpler, I think most likely you would come up empty. I don't see a simpler solution...
 
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  • #7
haruspex
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What if you considered just the horizontal motion and the viscous force acting on the horizontal component of the velocity?
Fine, but you first have to know or discover that it is ok to treat motion components separately for linear drag. For any other function of velocity, the drag due to the overall motion has to be found, then the component of interest taken.
 
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  • #8
Charles Link
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@Jahnavi Post 7 by @haruspex points out an important feature in the solution of this problem. The equations of motion ## \vec{F}=m \vec{a}=m \frac{d \vec{v}}{dt}=-mg \hat{z} -.2 \vec{v} ##, (with ## m=1 ##), are such that it made for a simple linear differential equation when the ## \hat{x} ## component is considered. We glossed over this in the subsequent discussion, but it is a key feature in the solution. ## \\ ## e.g. Had the viscous force been of the form ## \vec{F}_v=-Av^2 \frac{\vec{v}}{|\vec{v}|} =-A (v_x^2+v_y^2+v_z^2) \frac{\vec{v}}{|\vec{v}|} ##, (or any other functional form as @haruspex points out), it would have resulted in a non-linear differential equation, and compounding the difficulty, the components would not have separated like they do. Any solution would have been extremely difficult, and there would have been a function of the form ## f(v_x,v_z) ## in the ## \hat{x} ## component of the vector differential equation i.e. ## \frac{d v_x}{dt}=f(v_x,v_z) ##, etc. We would have had "coupled" differential equations instead of a single simple one to solve.
 
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Thanks !
 
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