Energy problem involving a projectile (tricky)

  • Thread starter delsaber8
  • Start date
  • #1
47
1

Homework Statement


A ball (0.45kg) is kicked at 32m/s, 45° above the horizontal. If the average force of air resistance is
1N, how far will it go?


Homework Equations


so I figure you would need energy to solve this question but I'm having trouble finding the total distance it travels, not just the horizontal distance but the parabolic-like distance. Am I on the right track? I have some of the things figured out but I'm finding it a little overwhelming.

The Attempt at a Solution


Loss to friction W=Fd W=(1)d
Ep=mgh=mgdy=(0.45)(9.8)dy for potential energy at maximum

Ek=1/2(0.45)(32)2 for kinetic energy at start

Y direction
Vo= 32 sin 45°= 22.63m/s
V=?
to find acceleration (this could be wrong) F=ma= 1=0.45a which means a=2.22 but the y component =1.57+g=11.37 down
dy=?
t=?

X direction
Vo= 22.63m/s
V=?
Did the same thing to find a here
a=-1.57
dx=?
t=?
 

Answers and Replies

  • #2
nasu
Gold Member
3,786
439
It may be trickier than you think.
I suppose the air resistance is in direction opposite to the velocity, so it changes direction.
You will have variable accelerations on both x and y directions.
What is the meaning of "average force" here? Is this form a book or you made it up?
 
  • #3
47
1
It may be trickier than you think.
I suppose the air resistance is in direction opposite to the velocity, so it changes direction.
You will have variable accelerations on both x and y directions.
What is the meaning of "average force" here? Is this form a book or you made it up?

This was made by my teacher, as for the air resistance I think it means that at any point the force of friction with the ball against the air is 1N (of course it may be more or it may be less), just so we don't have to deal with increasing resistance as the ball speeds up.
 
  • #4
nasu
Gold Member
3,786
439
And what will be the direction of this force? This matters too.
 
  • #5
906
88
...so we don't have to deal with increasing resistance as the ball speeds up.
You're right that you don't have to deal with increasing resistance with speed but, as already pointed out, you do have to worry about the variable direction of the resistance.

The only way I can think of solving this is using a numerical method.
 
  • #6
47
1
And what will be the direction of this force? This matters too.

But could you not just completely overlook that and just use the work formula W=F*d and since F is 1N we know that the amount of energy lost to friction (air resistance) is equal to the distance? And solve using conservation of energy instead, or am I missing something important?
 
  • #7
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
But could you not just completely overlook that and just use the work formula W=F*d and since F is 1N we know that the amount of energy lost to friction (air resistance) is equal to the distance? And solve using conservation of energy instead, or am I missing something important?
But you don't know the distance it travels, nor the time spent in the air. The path taken is roughly parabolic, not exactly parabolic.
 
  • #8
47
1
Exactly my problem, if there were no air resistance, one could probably use a quadratic equation to model its flight path. Any advice as to how I might tackle it?
 
  • #11
107
13
This would be pretty hard to solve using work and energy, considering you'd have to find the arc length of the trajectory (which isn't always an easy task) to know the distance over which your air resistance is applied.

Use Newton's 3rd law and the Kinematics, it'll be a lot easier.

You can see that $$a_{x} = -\frac {F_{D}}{m}$$ and $$a_{y} = - ( g + \frac {F_{D}}{m} )$$

These are found by comparing forces in each direction. Note ##F_{D}## is the drag force/air resistance.

So, plugging into the kinematics for the y direction we find $$t = \frac {v_{0}\sin \theta}{g+\frac {F_{D}}{m}}$$

Now we should multiply this time by 2 considering it is the time to get to the top of the trajectory, but we want the total time.

So, plugging into another one of the kinematics for the x direction: $$x = v_{0}\cos \theta t - \frac {F_{D}}{m}t^2$$

and using ##t = 2\frac {v_{0}\sin \theta}{g+\frac {F_{D}}{m}}##

You can simplify all of this into a decently clean equation for the total trajectory.

$$X_{traj} = \frac {v_{0}^2}{g+\frac {F_{D}}{m}} (\sin 2\theta - \frac {F_{D} - \cos2\theta}{mg+F_{D}})$$
 
  • #12
906
88
The above formula would be correct if F_D was the same in the x-direction and the y-direction. However, the problem strictly as interpreted gives the average value or magnitude as constant without specifying the direction. This makes the problem difficult to solve if not impossible using a direct method.

Perhaps the OP can clarify if the drag force F_D is supposed to be the same in each direction or if it is a constant magnitude whose direction is opposite to the velocity?
 
  • #13
906
88
Ok I actually found a projectile motion simulator(below) and I entered the values. The only thing I'm having trouble with is the drag coeffficent, because I'm not sure what to set it at to achieve an average air resistance of 1N?

http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html
That's a cute little program. The drag force it is using is almost certainly one that is proportional to velocity. The difference with your problem is that the magnitude in your problem is given as a constant. You might want to clarify this with your teacher.
 
  • #14
107
13
The above formula would be correct if F_D was the same in the x-direction and the y-direction. However, the problem strictly as interpreted gives the average value or magnitude as constant without specifying the direction. This makes the problem difficult to solve if not impossible using a direct method.

Perhaps the OP can clarify if the drag force F_D is supposed to be the same in each direction or if it is a constant magnitude whose direction is opposite to the velocity?

This is a rather intuitive principle of a consistent medium like air, and is a fair assumption that the force would have the same resisting value for both components of motion.

Also, for the program, if you leave the radius and drag force alone and merely enter in the values explicitly given in the question, there's roughly a 0.2% error between the experimental and calculated trajectory.

You're right though. This question is a bit vague on many values, considering drag force is not really an easy value to calculate.
 
  • #15
nasu
Gold Member
3,786
439
This is a rather intuitive principle of a consistent medium like air, and is a fair assumption that the force would have the same resisting value for both components of motion.
So even if there is no motion along some direction, there will be a drag force along that direction?
It does not seem very intuitive.
If you drop a ball, will it experience a lateral drag force?
Or if you launch it at 89 degrees upward, the drag force for the horizontal component (which is almost zero) will be the same as for the vertical motion?
 
  • #16
107
13
So even if there is no motion along some direction, there will be a drag force along that direction?
It does not seem very intuitive.
If you drop a ball, will it experience a lateral drag force?
Or if you launch it at 89 degrees upward, the drag force for the horizontal component (which is almost zero) will be the same as for the vertical motion?

Considering it is given purely as a constant force, then yes. In this equation, that's exactly what it is. If your drag force is a constant and there are no other forces in the horizontal component, your acceleration will always be $$a_{x} = -\frac {F_{D}}{m}$$

And summing up your forces in the y direction will give: $$ma_{y} = -mg - F_{D}$$
I will say you are right with some of the extreme cases. The equation I gave was under the assumption that the projectile stayed in motion for the entire time it was in the air. If the projectile's horizontal component is so small that the air resistance stops it before it even hits the ground, then you would calculate your trajectory using a time value of $$t = \frac {mv_{0}\cos \theta}{F_{D}}$$
 
  • #17
nasu
Gold Member
3,786
439
No, not even in the (confusing) text of the problem there is anything to grant the assumption that each component of the force has a constant value of 1N.
How do you calculate the "average" to get 1N with your assumptions of 1N on each direction?

And this is not considering the fact that the assumptions are unnatural.
At the top part of the trajectory the vertical motion is much slower than the horizontal. However you assume that the drag is the same.

I think is more "natural" to question the validity of the problem rather than trying to force reality to fit the problem.
 
  • #18
107
13
No, not even in the (confusing) text of the problem there is anything to grant the assumption that each component of the force has a constant value of 1N.
How do you calculate the "average" to get 1N with your assumptions of 1N on each direction?

And this is not considering the fact that the assumptions are unnatural.
At the top part of the trajectory the vertical motion is much slower than the horizontal. However you assume that the drag is the same.

I think is more "natural" to question the validity of the problem rather than trying to force reality to fit the problem.

The problem is unrealistic, but the equation fits the problem. If you somehow had a drag force that was constant (which is retarded, considering it's a function of the velocity normally) then you could use that equation and get the correct answer. So yes, obviously that equation wouldn't even work in the real world. We'd probably set it up as a first order linear differential equation.
 
  • #19
nasu
Gold Member
3,786
439
What kind of averaging (in your model) will give you an average drag force of 1N, in order to satisfy the problem?
Your total drag force is about 1.4 N all the time.
 
  • #20
47
1
The above formula would be correct if F_D was the same in the x-direction and the y-direction. However, the problem strictly as interpreted gives the average value or magnitude as constant without specifying the direction. This makes the problem difficult to solve if not impossible using a direct method.

Perhaps the OP can clarify if the drag force F_D is supposed to be the same in each direction or if it is a constant magnitude whose direction is opposite to the velocity?

I am afraid, I know only as much as is written in my original post, as I copied the question out exactly as it was written. It probably should be noted, that my teacher came up with the question which means he could have possibly made an error in his calculations, perhaps assuming it was a parabola, as unlikely as that would be for a teacher to make a simple error like that.
 
  • #21
906
88
...as unlikely as that would be for a teacher to make a simple error like that.

You would be surprised.
 
  • #22
47
1
You would be surprised.

Well, If and when I get the solution I will bring it here for everyone to look at and hopefully verify, however if he makes a blatant error like the one I mentioned earlier it shouldn't be too hard spotting it.
 
  • #23
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
If this were an exam question, then the examiner's aim is to test your understanding. One thing is certain---he can't justify to the College Board his awarding you marks if you do nothing. So how to make the best of this awkward situation, and give the examiner good reason for awarding you some marks, perhaps near full marks?

I suggest that you treat it as though drag can be approximated by a constant horizontal force, together with an equal-magnitude vertical drag force, and see what results. There is enough physics involved in such a scenario that your examiner will be able to appraise your understanding, and justify awarding you good marks. Faced with a question of a level of difficulty that seems beyond what should be expected, I think that's a reasonable compromise all 'round.
 
  • #24
47
1
Ok so I have been given a description of how to calculate this. My teacher claims that he calculated 1 second intervals, to find the way the components of force change during the flight. He also claims to have made an estimation for time it would take in his equation. I'm not really sure how this all ties together because I didn't see it. But by the way it looks I think integration might be necessary to solve. If anybody can make sense of this that would be great.
 
  • #25
906
88
He did it numerically.

Can you set up a spreadsheet?
 

Related Threads on Energy problem involving a projectile (tricky)

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
4
Views
4K
Replies
1
Views
2K
Replies
11
Views
874
  • Last Post
Replies
10
Views
3K
Replies
4
Views
12K
Replies
11
Views
10K
Replies
28
Views
3K
Top