# Projectile with air resistance

1. Mar 20, 2006

### 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??

2. Mar 20, 2006

### 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.

3. Mar 20, 2006

### 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??

4. Mar 20, 2006

### Hootenanny

Staff Emeritus
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.

5. Mar 20, 2006

### 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?

6. Mar 20, 2006

### 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

7. Mar 20, 2006

### Hootenanny

Staff Emeritus
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?

8. Mar 20, 2006

### nrqed

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

9. Mar 20, 2006

### 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.'

10. Mar 20, 2006

### Hootenanny

Staff Emeritus
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.

11. Mar 20, 2006

### 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?

12. Mar 20, 2006

### 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?

13. Mar 20, 2006

### Hootenanny

Staff Emeritus
I know, I've realised 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.

14. Mar 20, 2006

### nrqed

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

15. Mar 20, 2006

### 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.

16. Mar 20, 2006

### nrqed

Can you give us your v(t)?

17. Mar 20, 2006

### Hootenanny

Staff Emeritus
Yeah, I realised 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:

18. Mar 20, 2006

### Cyrus

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

19. Mar 20, 2006

### 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?

20. Mar 20, 2006

### Cyrus

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