Projectile with Air resistance problem

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Homework Help Overview

The problem involves a projectile experiencing air resistance that is proportional to its speed. The context includes determining the constant of proportionality using terminal velocity and formulating the equation of motion for a projectile thrown vertically upwards.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational force and air resistance, questioning the signs and dimensions of the forces involved.
  • Some participants attempt to derive the constant k and express the equation of motion, while others express confusion about the correct setup of forces and the implications of terminal velocity.
  • Questions arise regarding the direction of forces acting on the projectile and the correct interpretation of acceleration in the context of the problem.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have attempted to clarify their understanding of the forces and their signs, while others are still grappling with the dimensional consistency of their equations. Guidance has been offered regarding the need to reconsider the signs and units in their equations.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the physics concepts rather than simply arriving at a solution.

ZenchiT
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Homework Statement


The resistance to motion of a projectile, of mass m, is proportional to its speed. When falling vertically its terminal velocity is 19.6 m/s. The projectile is thrown vertically upwards with gravity acting downwards.
(a) Assume that the resistance to motion can be expressed as kv where v is the velocity andk is a constant. Determine k in terms of m using the terminal velocity value 19.6 m/s.
(b) Assume that the projectile is launched vertically upwards with a speed of 6 m/s. Write down the equation of motion which describes the resulting motion and find the expression
for the velocity, v, at time t.

Homework Equations

The Attempt at a Solution



a) I think I've got part (a) correct. I did g/19.6 = k, therefore constant k = 1/2
b) My biggest problem with this question is working out the resultant force. The way I would attempt this question is work out the resultant force and then use the equation F=ma. I would substitute a=dv/dt and then simply solve the differential equation by making v the subject. However I can't seem to work out what F would be.
I know that you would have to subtract mg (weight) from the projectile force.
My attempt would be
mkv-mg=ma
kv-g=dv/dt

Thank you so much for your help!
 
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ZenchiT said:
I know that you would have to subtract mg (weight) from the projectile force.
My attempt would be
mkv-mg=ma
kv-g=dv/dt
Is force of gravity upwards or downwards ? Does air resistance act upwards or downwards ?

You'll need to use a bit of calculus here.

Hope this helps.
 
The force of gravity would be downwards, so would mg be positive? And the air resistance would act upwards, so on the left handside you would have mkv+mg? And then you proceed to integrate it?

Thank you!
 
ZenchiT said:
The force of gravity would be downwards, so would mg be positive? And the air resistance would act upwards, so on the left handside you would have mkv+mg? And then you proceed to integrate it?

Thank you!
No, your final step is correct, but your reasoning is not. If velocity is upwards, and air resistance opposes it, will it act upwards or downwards ?
 
ZenchiT said:
a) I think I've got part (a) correct. I did g/19.6 = k, therefore constant k = 1/2
Afraid not. What are the units of kv? of g? Can you really equate g = kv?
b)
My attempt would be
mkv-mg=ma
kv-g=dv/dt
Again, your equation contains terms of unequal dimensions (units). Fix this first.
Then, reconsider the signs in your equation.
 
rude man said:
Afraid not. What are the units of kv? of g? Can you really equate g = kv?

Again, your equation contains terms of unequal dimensions (units). Fix this first.
Then, reconsider the signs in your equation.
I'm sorry I don't quite understand :/, if the dimensions are incorrect, what would I have to do?
 
Qwertywerty said:
No, your final step is correct, but your reasoning is not. If velocity is upwards, and air resistance opposes it, will it act upwards or downwards ?
Oh okay I think I get it! So if velocity is upwards then air resistance will act downwards, does that mean you have air resistance AND gravity opposing the velocity?
 
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ZenchiT said:
I'm sorry I don't quite understand :/, if the dimensions are incorrect, what would I have to do?
In your original question, you have said that kv is a force. Does g represent force or acceleration ?
ZenchiT said:
Oh okay I think I get it! So if velocity is upwards then air resistance will act downwards, does that mean you have air resistance AND gravity opposing the velocity?
Yes, that's correct !
 
  • #10
Qwertywerty said:
In your original question, you have said that kv is a force. Does g represent force or acceleration ?
Yes, that's correct !
ZenchiT said:
Oh okay I think I get it! So if velocity is upwards then air resistance will act downwards, does that mean you have air resistance AND gravity opposing the velocity?
No, that's incorrect.
If I drop something does gravity act to oppose velocity? Try jumping off a roof! :smile:
 
  • #11
ZenchiT said:
I'm sorry I don't quite understand :/, if the dimensions are incorrect, what would I have to do?
Why not try Newton: sum of forces = mass times acceleration? And you need to decide what the signs of kv and mg are.
 
  • #12
rude man said:
Why not try Newton: sum of forces = mass times acceleration? And you need to decide what the signs of kv and mg are.
Haha! Now I'm super confused! I'm getting two different answers :/
 
  • #13
I have had another attempt at question a). mg - mkv = ma, therefore when terminal velocity =19.6, k in terms of m is equal to, k = (m(g-a))/19.6

Is that correct?
 
  • #14
ZenchiT said:
Haha! Now I'm super confused! I'm getting two different answers :/
On PF, one of the things you have to decide is to whom to listen!
 
  • #15
ZenchiT said:
I have had another attempt at question a). mg - mkv = ma. Is that correct?
No, you are still mixing units.
, therefore when terminal velocity =19.6, k in terms of m is equal to, k = (m(g-a))/19.6
what is a ?
 
  • #16
rude man said:
No, you are still mixing units.

what is a ?
a is acceleration
 
  • #17
ZenchiT said:
a is acceleration
What acceleration? Accel varies from the moment the projectile is fired in the UP direction until terminal velocity is reached in the DOWN direction .. so what number do you plan to ascribe to a?
 
  • #18
Ahh okay, would I be able to write it as dv/dt, and then solve to get velocity?
 
  • #19
Can someone please run me through this, I'm more confused than ever! Sorry to be a pain :/

Thank you for all the support so far
 
  • #20
rude man said:
Afraid not. What are the units of kv? of g? Can you really equate g = kv?

Again, your equation contains terms of unequal dimensions (units). Fix this first.
Then, reconsider the signs in your equation.
In terms of unequal dimensions, I believe the dimensions are correct? Because the equation I have is v= (g/k)(1- e^-kt). where g is gravitational acceleration and k is a constant. Terminal velocity is when t tends to infinity. If t tends to infinity then e^-kt tends to 0. Therefore v = (g/k). Why isn't that correct?
 
  • #21
ZenchiT said:
Can someone please run me through this, I'm more confused than ever! Sorry to be a pain :/

Thank you for all the support so far
With regard to signs...
The question as set has the initial velocity as upwards. It does not specify whether the equation you are to provide is to cover only that phase or whether it should be valid for all time. Think about the directions of gravity and drag in the upward and downward cases. What is different? How can you deal with that?

With regard to dimensions...
You are given that air resistance is kv. Is resistance a force or an acceleration?
 
  • #22
ZenchiT said:
In terms of unequal dimensions, I believe the dimensions are correct? Because the equation I have is v= (g/k)(1- e^-kt). where g is gravitational acceleration and k is a constant. Terminal velocity is when t tends to infinity. If t tends to infinity then e^-kt tends to 0. Therefore v = (g/k). Why isn't that correct?
What are the units of kv? That is your first task now.
v = g/k is not a dimensionally consistent equation. And the exponent has to be dimensionless. kt is not dimensionless.
 
  • #23
I believe I've finally got the answer to part a)!
F=ma and F=mg (gravity acting downwards) - kv (resistive force)
Therefore...
mg-kv=ma
Rearranging, where v=19.6
k= m(g-a)/19.6

Is that correct?!
 
  • #24
Please tell me that's correct!
 
  • #25
ZenchiT said:
I believe I've finally got the answer to part a)!
F=ma and F=mg (gravity acting downwards) - kv (resistive force)
Therefore...
mg-kv=ma
Rearranging, where v=19.6
k= m(g-a)/19.6

Is that correct?!
Almost right. You are told the teminal velocity. What value does that imply for a?
 
  • #26
haruspex said:
Almost right. You are told the terminal velocity. What value does that imply for a?

That it is equal to 0?! So we have 9.8m/19.6=k, therefore k=m/2?!

Thank you very much for your help? Any chance I can get the solution for part b), need it really urgently :/
 
  • #27
ZenchiT said:
That it is equal to 0?! So we have 9.8m/19.6=k, therefore k=m/2?!

Thank you very much for your help? Any chance I can get the solution for part b), need it really urgently :/
You cannot write k = m/2 because that is only true in a specific system of units. When you plug in the values for g and v, you should plug in the units too. There will be some cancellattion, yielding a set of units to apply to m/2. (Just treat the units, like kg, m, s as though they are algebraic variables.)

For part b, no we do not simply give solutions, no matter how urgent. You should now be able to write the equation relating forces and acceleration.
 
  • #28
rude man said:
No, that's incorrect.
If I drop something does gravity act to oppose velocity? Try jumping off a roof! :smile:
We are only discussing the case in which the ball goes up.
 
  • #29
Qwertywerty said:
We are only discussing the case in which the ball goes up.
Not exactly. The terminal velocity is given, and a deduction is to be made from it. That implies considering the downward stage.
 
  • #30
haruspex said:
Not exactly. The terminal velocity is given, and a deduction is to be made from it. That implies considering the downward stage.
I was discussing part b.
 

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