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[Quick question] A bird's fall.

  1. Mar 28, 2008 #1
    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?
  2. jcsd
  3. Mar 28, 2008 #2
    ... you are not precise enough in your description. Does it also retract its wings?

    Technically the bird is falling the entire time that it is in the air. Think of it's net motion (flying) as a summation of 1. falling, 2. pushing down the air below it. If it stops flapping, you eliminate 2, and then you are just simply falling.

    so i guess to answer your question, unless i misunderstood it, yes it falls immediately, according to the parabolic motion, subject to gravitational acceleration g, just like anything else
    Last edited: Mar 28, 2008
  4. Mar 28, 2008 #3


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    is it an unladen swallow?

    and if so is it an african or european swallow?

    either way, i dont know.
  5. Mar 28, 2008 #4
    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?
  6. Mar 28, 2008 #5
    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.
  7. Mar 28, 2008 #6
    Okay thanks.
  8. Mar 28, 2008 #7


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    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.
  9. Mar 28, 2008 #8
    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?
  10. Mar 28, 2008 #9
    yes but even then it's moving up, then identically 0 for just single point, and then moving down. There is no range of time at which the ball has speed 0.
    i.e. its not like the ball is thrown up, then it sits up at rest for 1 second and then it falls down.
  11. Mar 28, 2008 #10


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    Other than aerodynamic drag, the bird is in free fall when it's not flapping it's wings. There are some local small birds that do exactly what you describe, they flap their wings for about 1 second, then fold them up for another 1 second, repeating this cycle while flying. Sometimes the cycle is bit faster.
  12. Mar 28, 2008 #11


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    I agree, but that single point is still a point. Unless time turns out to be quantized, that point is immeasurably small. Nevertheless, it exists. For all practical purposes, your original answer was correct; I just like to cover all of the bases, even if they're theoretical.
  13. Mar 28, 2008 #12

    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 theres 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 small velocities like 0.00002m/s. So we can measure the time between the point where the ball is travelling up at 0.00002m/s and travelling down at 0.00002m/s.

    Does this make sense?
    Last edited: Mar 28, 2008
  14. Mar 28, 2008 #13
    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 thats 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?
  15. Mar 29, 2008 #14
    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?
    Last edited: Mar 29, 2008
  16. Apr 2, 2008 #15
    Oh wow yes, it makes sense. Thank you.
  17. Apr 6, 2008 #16
    It depends on how small your units when measuring time. In a perfect model, there it a POINT in time when the ball is motionless, but this does not take up any AMOUNT of time.

    Sorry i'm bad at explaining things.
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