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wind resistance and x displacement

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deleted783484


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This is a question from TBR translational motion Phase II

If all 4 objects were launched from a cannon mounted at a 45 degree angle what would be observed?
I. Object I would have the greatest range
II. Object IV would have the lowest velocity upon impact of the ground
III. The object experiences a greater displacement in the x direction in the first half of the flight than the second half of the flight

Given in the passage:

- Time of flight
I: 2.13s, II 2.14s, III 3.87s , IV 2.21s
- Wind resistance is proportional to velocity when the wind resistance is strong enough it can negate any acceleration and result in constant velocity

Answer is I and III

I understand why I is correct and II is incorrect but III seems strange to me. TBR says " Because of wind resistance opposing the direction of velocity it is constantly slowing the object down during its flight. this means that the object has greatest x direction velocity initially but that value is decreasing with time"
This statement is a bit confusing does it mean that the wind resistance is slowing it down more and more and making it displace less and why would that be the case?

Thanks!
 

aldol16

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Yes, it's saying that the drag force exerted by the wind resistance will oppose the direction of motion, thereby causing a deceleration in the x direction. In other words, drag force = -m*a = -m*dv/dt. Therefore, velocity in the x direction will always be decreasing due to this force. Thus, in the first half of the flight, it will have a greater average velocity than in the latter half of the flight. Greater average velocity = greater displacement given the same time.

But I'm not sure I understand the first part. That, for four objects launched at the same angle with the same force, the object that has the longest time in flight should have the longest range.
 
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deleted783484

Yes, it's saying that the drag force exerted by the wind resistance will oppose the direction of motion, thereby causing a deceleration in the x direction. In other words, drag force = -m*a = -m*dv/dt. Therefore, velocity in the x direction will always be decreasing due to this force. Thus, in the first half of the flight, it will have a greater average velocity than in the latter half of the flight. Greater average velocity = greater displacement given the same time.

But I'm not sure I understand the first part. That, for four objects launched at the same angle with the same force, the object that has the longest time in flight should have the longest range.
But how do we know that the drag force is decrease the velocity more and more and not decelerating at the same magnitude and so decreasing the velocity the same throughout?

For the first statement object 1's flight time was shorter so there was less drag force keeping it in air and stopping it from flying so since there is less drag force opposing the velocity (in the x direction) it will have the highest velocity and fly the farthest.
 

aldol16

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But how do we know that the drag force is decrease the velocity more and more and not decelerating at the same magnitude and so decreasing the velocity the same throughout?

It doesn't matter whichever way you look at it - velocity is decreasing throughout and therefore the second half of the flight will see less displacement. Imagine this. A uniform drag force acts on the object of -1 m/s^2. The initial x-velocity is 10 m/s. After 1 s, it's 9 m/s. After 2 s, it's 8 m/s. And so on. So let's say it goes all the way down to 5 m/s during the first half of flight and then to 0 m/s at the end of the flight. x-displacement is the integral of that velocity function. In other words, after the first second, it has traveled 10 m. After the second second, it has traveled 19 m. If you do the math in your head, you'll see that because it's inevitably slower during the second half of flight, the displacement must be less.

For the first statement object 1's flight time was shorter so there was less drag force keeping it in air and stopping it from flying so since there is less drag force opposing the velocity (in the x direction) it will have the highest velocity and fly the farthest.

I'm not sure that's an adequate explanation. Sure, it's flight time is shorter but your analysis would only be true if these objects all had the same mass. Do they?
 
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deleted783484

It doesn't matter whichever way you look at it - velocity is decreasing throughout and therefore the second half of the flight will see less displacement. Imagine this. A uniform drag force acts on the object of -1 m/s^2. The initial x-velocity is 10 m/s. After 1 s, it's 9 m/s. After 2 s, it's 8 m/s. And so on. So let's say it goes all the way down to 5 m/s during the first half of flight and then to 0 m/s at the end of the flight. x-displacement is the integral of that velocity function. In other words, after the first second, it has traveled 10 m. After the second second, it has traveled 19 m. If you do the math in your head, you'll see that because it's inevitably slower during the second half of flight, the displacement must be less.



I'm not sure that's an adequate explanation. Sure, it's flight time is shorter but your analysis would only be true if these objects all had the same mass. Do they?

They don't specify anything about the mass of the objects but they do say that the objects are different. Also in the passage they said that the greater the flight time the greater the wind resistance and consequently lower the terminal velocity.
 
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