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Basic physics – mechanics (it is not homework)

  1. Aug 5, 2008 #1
    Hi everyone,

    I would like to ask somebody here that could better understand than me of the following example from a book Physics – Principles with Applications, 3rd ed. by Douglas C. Giancoli, Prentice Hall, page 33. Here is the example http://img231.imageshack.us/img231/8103/fleajb0.jpg" [Broken]

    Example 2-12
    The jump of a flea can be analyzed using slow-motion photography. The motion can be separated into two parts. The first is the “push off,” lasting about 10-3 second, during which the flea’s legs push against the ground and accelerate it to a speed of about 1.0 m/s. The second part is the flight of the flea into the air subject only to gravity http://img231.imageshack.us/img231/8103/fleajb0.jpg" [Broken] Calculate:

    (a) the acceleration of the flea during push off expressed as a multiple of g, the acceleration of gravity;
    (b) the distance above the ground the flea reaches during push off; and
    (c) the height the flea should reach during the second period when its acceleration is that of gravity. Assume the jump is vertical.

    Let us take the upward direction as positive, with y0 = 0.

    (a) During the push off, the acceleration is

    [tex]a = \frac{v - v_{0}}{t} = \frac{1.0 \ m/s - 0}{10^{-3} \ s} = 1000 \ m/s^2.[/tex]

    The magnitude of the acceleration is (1000 m/s2)/(9.80 m/s2) [tex]\approx[/tex] 100 times the acceleration of gravity. Thus a [tex]\approx[/tex] 100 g.

    (b) We use equation [tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2 ,[/tex] replacing x with y:

    [tex]y = \frac{1}{2}at^{2} = 0.5(10^{3} \ m/s^{2})(10^{-3} \ s)^{2}
    = 0.5 * 10^{-3} \ m = 0.05 \ cm[/tex]

    (c) Now the acceleration is solely due to gravity, so a = -9.80 m/s2 and the initial velocity for this second period is 1.0 m/s. We use equation [tex]v^{2} = v_{0}^{2} + 2a(x - x_{0}),[/tex] with v = 0 (we are considering the highest point):

    [tex]y = \frac{v^{2} - v_{0}^{2}}{2a} = \frac{0 - (1.0 \ m/s)^{2}}{2(-9.80 \ m/s^{2})} = 0.05 \ m = 5 \ cm.[/tex]

    Photographs indicate that the flea jumps only about two-thirds this high. Can you guess why our calculation is off?

    All of the example and solutions was just mentioned above. So can anyone tell me what the author means by the “calculation is off”? This problem is so difficult for me and makes me can’t do the exercise in this chapter. I need to understand this example of problem first.

    Thank you very much on your help for this
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 5, 2008 #2


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    Can you think of any factors the would have affected the height of the flea's jump and which you didn't take into account?
  4. Aug 5, 2008 #3
    Hi Hootenanny,

    Thank you for your reply.

    I am sorry but…..

    I really don’t know which factor is it!

    All of the factor

    [tex]v_{0}, v, a / g, y_{0}, y[/tex]

    is already known and obtained unless its time or

  5. Aug 5, 2008 #4
    Don't consider mathematical factors, consider physical factors. Remember that the mathematics is only a model, and the flea is in the real world.
  6. Aug 6, 2008 #5


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    As Varnick has already said, you should be looking at physical factors. For example, are there any forces that would act on the flea in real life that your model hasn't taken into account?
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