Projectile Motion with Linear Air Resistance

In summary, unscientific tried to solve part (a) by solving the differential equations in the x and y directions, but for the approximation maximum range part he's slightly confused, since he can't seem to find the xmax using the given information. He then tries to approximate it with a linear equation and ends up with: 2VoUo/g = x(1+2kx/3Uo). This gives him the xmax for a given angle.
  • #1
unscientific
1,734
13

Homework Statement


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The Attempt at a Solution



I've solved part (a) by solving the differential equations in the x and y directions, but for the approximation maximum range part I'm slightly confused:

1. What does it mean by "assuming the correction with k ≠ 0 is small? I tried to expand ln (1 + y) = y - y2/2 + ... to the first order but that doesn't give me a result..(unless you consider x = 0 as a result)

2. Why are there 2 maximum ranges here? The first bit is about finding the range right? Why do they call it maximum range when it is just the range at any given angle?
I know the last part involves finding the angle that gives the maximum range, which is truly the maximum range, and not as described as above..
 
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  • #2
I think you can use ##ln(1-x) \approx -x## as approximation. This gives a linear equation, which should have two solutions (one is the trivial x=0).
Edit: Hmm ok, you need more terms.

2. Why are there 2 maximum ranges here?
Where do you get 2 maximum ranges? The maximum range is the range at the optimal angle.
 
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  • #3
Hi unscientific
1) As you suspected, k being small enough means it is small enough that we know that |kx/u0|<1 so that you can do what you just did, but try to go to the third order instead
2) in the first part the velocity is given, for this given you could find an xmax, there you can also try to find a xmax for a given angle (u0 and v0 would change in your first equation)

Cheers...
 
  • #4
Retain the terms till the third order, inclusive.
 
  • #5
mfb said:
I think you can use ##ln(1-x) \approx -x## as approximation. This gives a linear equation, which should have two solutions (one is the trivial x=0).
Edit: Hmm ok, you need more terms.


Where do you get 2 maximum ranges? The maximum range is the range at the optimal angle.


When i tried approximating the ln function to -x, you factor out the x and it doesn't give any result...

The question keeps mentioning "maximum range", when in the first part it is simply referring to the horizontal range. Only in the last part are we supposed to find what angle gives the maximum range, and what the maximum range is.
 
  • #6
Hi again unscientific
in the first part, by 'maximum range' xmax, what is meant is where is the projectile going to hit the ground.
this is in this way that it is 'xmax' if you shot from a hill, it could go farther horizontally, but that's not the point.
there you get a formula.
ln(1-blabla)=ln(1+(-blabla)) and you know that blabla is small so you can expand it with your formula, do it to the third order and it should lead you to the answer

In the second part, we talk about having the greatest xmax possible (that is the real and only maximum range as stated by mfb) and that will depend on the angle at which you shoot your projectile. so this is what this second part is about.

Cheers...
 
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  • #7
I hate to bump an old thread, but it might be slightly better than creating a new one to ask what is essentially the same question.

I expanded the logarithmic function to third order in k, and ended up with:

2VoUo/g = x(1+2kx/3Uo)

after dividing through by x and rearranging a little which doesn't seem to simplify to the answer they want. Do we need to make another approximation somewhere along the line?

ETA: Got it. Needed to use Binomial expansion twice. Feel deeply embarrassed.
 
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Related to Projectile Motion with Linear Air Resistance

1. What is projectile motion with linear air resistance?

Projectile motion with linear air resistance is a type of motion in which an object is launched into the air and moves along a curved path due to the force of gravity while also experiencing resistance from the surrounding air. This type of motion is commonly seen in activities such as throwing a ball or shooting a projectile.

2. How does air resistance affect projectile motion?

Air resistance, also known as drag, acts in the opposite direction to the motion of the object and can slow it down. This means that a projectile with air resistance will not follow a perfect parabolic path, as it would without air resistance. Instead, it will have a steeper trajectory and a shorter range.

3. What factors affect the amount of air resistance on a projectile?

The amount of air resistance on a projectile depends on several factors, including the object's shape, size, and speed. A larger or more streamlined object will experience less air resistance, while a smaller or more irregularly shaped object will experience more. Additionally, the faster an object is moving, the greater the air resistance it will experience.

4. How is the motion of a projectile with air resistance calculated?

The motion of a projectile with air resistance can be calculated using mathematical equations, such as the projectile motion equations for velocity and displacement, along with the drag force equation. These equations take into account the object's initial velocity, angle of launch, mass, air density, and other factors to determine its trajectory.

5. Can air resistance be ignored in projectile motion calculations?

In most cases, air resistance cannot be ignored in projectile motion calculations as it significantly affects the motion of the object. However, in situations where the object is small, slow-moving, or in a highly controlled environment, the effects of air resistance may be negligible and can be ignored for simplicity.

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