# Homework Help: Is there a quicker way to solve this projectile problem?

1. May 17, 2018 at 9:39 PM

### Jahnavi

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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 .

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 .

2. May 17, 2018 at 9:56 PM

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?

Last edited: May 17, 2018 at 10:29 PM
3. May 18, 2018 at 12:28 AM

### Jahnavi

I did the same

Yes !

4. May 18, 2018 at 12:31 AM

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...

Last edited: May 18, 2018 at 12:46 AM
5. May 18, 2018 at 1:00 AM

### Jahnavi

I can see , you are trying to make me feel better .

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. May 18, 2018 at 1:05 AM

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...

7. May 18, 2018 at 6:17 AM

### haruspex

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.

8. May 18, 2018 at 9:14 AM

@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.

Last edited: May 18, 2018 at 9:53 AM
9. May 18, 2018 at 10:58 AM

Thanks !