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b-pipe
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If a bird is in the air and for whatever reason it stops flapping it's wing, it'll fall back down. Is there a lag time between the stopping of the bird's wing and the fall? If there is, how long is it?
b-pipe said:Okay I may have used a wrong analogy. I throw a ball into the air, the ball reaches it's peak and starts to fall, What is the time between the ball reaching it's peak and the ball falling?
Okay thanks.Mephisto said:yes. it's exactly 0. the ball will not be suspended in air for any amount of finite time, if that is what you are asking.
Can you explain further? Because this kind of contradicts what Mephisto said. If there is a brief transition how can I find out the time lag?Danger said:One caveat to that would be if you can manage to throw the ball absolutely straight up with no wind or spin, it would very briefly come to a halt during the transition from 'up' to 'down'. Anything else would result in a parabola of some sort.
Danger said:One caveat to that would be if you can manage to throw the ball absolutely straight up with no wind or spin, it would very briefly come to a halt during the transition from 'up' to 'down'. Anything else would result in a parabola of some sort.
Mephisto said:yes but even then it's moving up, then identically 0 for just single point, and then moving down.
The number line example really put it into perspective and you really hit the nail on the head with "appear motionless to the human eye" part because that's what I had in mind. But I'm confused about how we can measure the time between the point where the ball is traveling up at 0.002 m/s and traveling down at 0.002m/s?grmnsplx said:b-pipe,
I think I see what you're asking. It's a fair question.
The ball reaches it's peak when the upward velocity becomes zero, right?
Note that the function describing its velocity (first upward very quickly, then slowing down, velocity becomes zero, starts falling slowly, then faster and faster...) is continuous.
Also note that the downward force of gravity is constant on the bird/ball. This means that, since there are no other forces acting on the ball, the acceleration (downward) is constant.
So mathematically, both of there's functions are continuous, and the derivative is differentiable at every point. So we have nice smooth functions to play with.
My point is that the ball remains motionless in the air (i.e. velocity = 0) only for an instant. When I say an instant I mean an arbitrarily small bit of time. So small we can't measure it. If you throw a ball in the air it will NEVER reach its peak and then stay there for 5 minutes and then decide to fall.
So, "What is the time between the ball reaching it's peak and the ball falling?" It's essentially zero.
Is this making sense? Let me continue...
Look at the number line. We can ask " what is the 'distance' between 3 and 5?" Easy. It's 2. take 0 and 1. the 'distance' is 1.
Your question is like asking "what is the distance between 0 and the next positive number?" "1!" No wait, not 1, there are more numbers in there. "1/2?" No, many more numbers are between 1/2 and 0. If you give me any positive number that you think is close to zero, i can find one that's closer. Can you see how small these numbers would get?
As I said before, we can't measure the length of time during which the velocity of the ball was 0. The same is true for any velocity in this case. For how long was the ball traveling at 1m/s? Only for an instant, because the next instant the ball was either going a bit faster (downwards) or slower (going up).
What we can measure easily is the the time between two velocities, such as the time it takes the ball to go from the ground at the initial velocity (like 10m/s) and it's peak (0m/s). Or in your question maybe we could define "appear motionless to the human eye" as really velocities like 0.002m/s. So we can measure the time between the point where the ball is traveling up at 0.002m/s and traveling down at 0.002m/s.
Does this make sense?
Oh wow yes, it makes sense. Thank you.grmnsplx said:OK. So we know that velocity is equal to acceleration by time, correct?
[tex] v = at [/tex] Since we are talking about acceleration due to gravity I should use the term g where g = -9.8m/s^2
[tex] v = gt [/tex]
Solving for t we have:
[tex] t = \frac{v}{g} [/tex]
Now what we want to know is the length of time, [tex] \triangle t = t_{2} - t_{1} [/tex], it takes to go from one velocity to another, [tex] (v_{1}, v_{2})[/tex].
[tex] \triangle t = \frac{v_{2}}{g} - \frac{v_{1}}{g}[/tex]
let's take really small numbers for velocity like 0.002m/s, and to make calculation easy let's use g=-10m/s^2. it's close enough.
[tex] \triangle t = \frac{-0.002m/s}{-10m/s^2} - \frac{0.002m/s}{-10m/s^2}[/tex]
[tex] \triangle t = 0.0002s - (-0.0002s) [/tex]
[tex] \triangle t = 0.0004s [/tex]
Notice that v2 is negative because it is going down.
Does this make sense?
b-pipe said:Can you explain further? Because this kind of contradicts what Mephisto said. If there is a brief transition how can I find out the time lag?
A bird can fall due to various reasons such as injury, exhaustion, collision with an object, or being caught in a strong wind.
Yes, birds can get injured from falling, just like any other animal or human. The severity of the injury depends on the height of the fall and the impact on the bird's body.
Yes, some birds can survive a fall depending on the height and surface they land on. Smaller birds have a better chance of survival compared to larger birds due to their lighter weight and ability to flap their wings to slow down their descent.
Birds spread their wings when falling to increase air resistance and slow down their descent. This helps them to land safely and avoid injuries.
The highest recorded fall by a bird that survived was a mallard duck that fell from a height of 4,000 feet and survived. However, the average height for a bird to survive a fall is around 50-100 feet.