Projectile with air resistance

[Jess1986]

A paticle of mass m is projected vertically upwards with speed U. The air resistance produces a retarding force mkv^2 , where k constant and v the speed of the particle. Find the greatest height reached by the particle. Determine the speed W with which the particle will return to the point of projection.

I have done the first part of this question using a=-g-kv^2
I found the greatest height reached to be (U^2(g-kv^2)) / 2g^2 i think i have done this right!

I am unsure how i then find the speed W?? Do i take downward motion as i new problem, so that when the particle is at greatest height t would=0??
Please help!

 

[vaishakh]

You can start with the equation v = 0 at beginnig and a = g + kv^2.
I think you used Calculus to solve your earlier question. Solve this question again using the same analysis.

 

[Jess1986]

Having real trouble finding a value for W for this question. As soon as i bring in the height value that i calculated above it has got very complicated with many U, g, k, v and squares! Could anybody give me a hand please please??

 

[Hootenanny]

I found the greatest height reached to be (U^2(g-kv^2)) / 2g^2 i think i have done this right!

I think you need to re-think this. Use the kinematic eqaution $v^2 = U^2 + 2as$. Your accleration you calculateed, $a = -(g + kv^{2})$ is correct.

 

[Jess1986]

ok thanks, should i need to should the derivation of this equation?
i have found the height reached now to be (u^2)/(2(g+kv^2)). This looks better i think?

 

[Jess1986]

how should i find the speed W? Using the same formula again gives W=U which i don't think is correct. Any suggestions? Thanks

 

[Hootenanny]

i have found the height reached now to be (u^2)/(2(g+kv^2)). This looks better i think?

Looks good to me

I think you may be able to do this question without calculus depends on what type of answer is required. Does it specify what terms W should be given in?

 

[nrqed]

I think you need to re-think this. Use the kinematic eqaution $v^2 = U^2 + 2as$. Your accleration you calculateed, $a = -(g + kv^{2})$ is correct.

Unless I am mistaken, the above equation (v^2 = U^2 + 2 a s) is valid only for constant acceleration so it can't be used here. Am I missing something?

Pat

 

[Jess1986]

No, the question just says 'determine the speed W with which the particles will return to the point of projection. Confirm that the dimensions of the result have units of speed.'

 

[Hootenanny]

Unless I am mistaken, the above equation (v^2 = U^2 + 2 a s) is valid only for constant acceleration so it can't be used here. Am I missing something?

Pat

Ahh, I forgot about the $v^2$, you are quite right Pat, thanks for catching us, before we went too far

Jess:
So you will need to integrate you expression for a in terms of v to find an expression expression for velocity and then again to find an expression for s.

 

[Jess1986]

that is what i did the first time and
I found the greatest height reached to be (U^2 (g-kv^2)) / 2g^2 do you agree with this?

 

[Cyrus]

Hootennany, how come you used:

$v^2 = U^2 + 2as$

which is an equation of constant acceleration, when the problem specifies a nonlinear acceleration?

 

[Hootenanny]

I know, I've realized that now, it was a BIG mistake on my part. You have to use calculus to solve this question.

Jess: Yes you equation is correct from what I can see.

 

[nrqed]

Unfortunatly no. Kinetmatic equations are only valid when there is uniform accleration, so, we can't use them in this instance. However, you greatest height reached is correct. IF we use the work-energy theorm;

$$\frac{1}{2}mu^2 = mgh + kmv^{2}h$$

Which can be manipulated into your formula. I don't quite know why we got the same answer.

Sorry to butt in again
But this is incorrect. This again assumes a constant acceleration (because the work is the integral of the force times the distance...It is equal to the force times the distance only for a constant force but the drag force here is not a constant force).

One has to integrate the acceleration equation to find v(t), I am afraid. maybe th eoriginal poster could give his answer for v(t)?

Pat

 

[Jess1986]

ok, i am slightly confused. so is the greatest height reached which i calculated correct? i need to use integration techniques, i have not studied work-energy principles for this course and so could not really include them.

 

[nrqed]

ok, i am slightly confused. so is the greatest height reached which i calculated correct? i need to use integration techniques, i have not studied work-energy principles for this course and so could not really include them.

Can you give us your v(t)?

 

[Hootenanny]

Sorry to butt in again
But this is incorrect. This again assumes a constant acceleration (because the work is the integral of the force times the distance...It is equal to the force times the distance only for a constant force but the drag force here is not a constant force).

One has to integrate the acceleration equation to find v(t), I am afraid. maybe th eoriginal poster could give his answer for v(t)?

Pat

Yeah, I realized it right after I typed it, that's why I deleted it. It's been a hectic couple of days, I think I need a break...:uhh:

 

[Cyrus]

To me, this solution should involve an iterrative set of equations. Maybe I am missing something.

 

[Jess1986]

when looking for greatest height reached i found

a= -g-kv^2
v= -gt-(kv^2)t + U
x= -0.5gt^2 - 0.5(kv^2)t^2 + Ut

i think these are all correct. i then found the time when v=0 which will be at the greatest height. This is t=u/g
i substituted this value for t into equation for x and got
x= (U^2(g-kv^2)) / 2g^2
what do you think?

 

[Cyrus]

No, a(t) and v(t) are functions of t. Your integration is not right.

 

[Hootenanny]

Yes, my intial thought on looking at the intergrals suggested an iterrative solution, but I discarded it because I thought I would be able to obtain an 'accurate' answer. Obviously not...

 

[Jess1986]

I do not understand. Why are they not correct?

 

[Cyrus]

a and v are functions of t. You integrated and just plopped a t infront of the v^2. That is not correct. (I may have a way to go about it, but I have to study right now. If no one gives you an answer, Ill try to post some thoughts later tonight.)

 

[nrqed]

To me, this solution should involve an iterrative set of equations. Maybe I am missing something.

That's what I thought too (I am not sure if a numerical solution is allowed here...) I am rusty in diff eqs so I could not see right away the closed form solution but I might be missing something simple..

Pat

People are poor, because society has denied them the real social & economic base to grow on.

And some people are poor because they don't want to put in the efforts to improve their situation...

 

[Jess1986]

Could you explain to me how the integration should be done correctly? Thankyou

 

[nrqed]

Could you explain to me how the integration should be done correctly? Thankyou

can you tell us a bit about your background? Have you learned differential equations? Are you allowed to use numerical integration? This will guide us in our explanations.

Pat

 

[Jess1986]

i am currently a 1st year undergraduate. I am taking this mechanics module at uni. I have previously done differential equations. I can use integration yes, but I am not sure what you mean by 'numerical' integration. Sorry if that's a stupid question!

 

[nrqed]

i am currently a 1st year undergraduate. I am taking this mechanics module at uni. I have previously done differential equations. I can use integration yes, but I am not sure what you mean by 'numerical' integration. Sorry if that's a stupid question!

It's noty a stupid question at all.. It's basically how to use a computer to solve diff equations numerically. For example, one could use Excel to solve your problem within a certain precision.

However, without that, what you have to do is to solve the differential equation

${dv \over dt} = - g - k v(t)^2$.

Let me ask something: are you sure the drag force is of the form - k v^2? (I am asking just in case...I thought that for small velocities there was also a linear form, -k v and in that case the solution would be easy).

Pat

 

[Jess1986]

Thanks.
The retarding force is definitely mkv^2 .
I have not used excel to solve problems before. How did you get to that differential equation?

 

[Jess1986]

oh i see now. my differential equation skills are not too great. So i now integrate both sides?

 

[nrqed]

oh i see now. my differential equation skills are not too great. So i now integrate both sides?

Sorry, I had left a factor of "m" on the left side which should not have been there.

You can't integrate since v(t) is an unknown function. The fact that it is squared makes this a nonlinear equation which is tough. I am trying to guess the solution but I don't see it right away, sorry.
(maybe Maple could give you the answer).

 

[Jess1986]

thanks, you have still helped me get on the right track. once i find v do you know where i should go from there?

 

[nrqed]

thanks, you have still helped me get on the right track. once i find v do you know where i should go from there?

In principle, here is what you have to do:

Sokve for v(t) with the initial condition v(0) = U.

set v(t)= 0 to find the time to get to the highest point.

Integrate v(t) to find y(t) (with the initial condition y(0)= 0). Plug in the time from the first part to find the max height reached.

Now start again, but now solving dv/dt = +g - k v^2 with the initial condition v(o) = 0.

Integrate v(t) to find y(t) with the initial condition being the max height.

Find the time at which y is zero again.

Plug back that time in the formula for v(t) on the way down.

It's that easy

 

[Jess1986]

thanks very much for all your help. il give solving the equation in maple a go tomorrow and see how it goes!