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Calculating air resistance

  1. Apr 16, 2005 #1

    While searching for a solution to a particular problem, I stumbled upon this forum; any help would be greatly appreciated.

    We're currently working on energy and power, and in spite of my past success with such problems, I couldn't solve this one:

    A volleyball is dropped from a height of 10m, but hits the ground with a velocity of only 10m/s.

    Calculate the quantity of work done by air resistance on the ball.

    Now, as it turns out, our teacher told us that the problem could be solved without the mass, however, he also said that it could prove to be complicated, therefore, he gave us the mass of the ball (500 g). He also gave us the answer, which is 24J.

    Solving it with the mass is obviously simple: I determined the potential energy and the kinetic energy; if there was a difference between both of them (technically both should be the same due to the fact that the given speed is just before it hits the ground), it was related to the air resistance.

    Ep = .5kg x 9.8N x 10
    Ep = 49J

    Ek = (.5kg x 10^2)/2
    Ek = 25J

    49-25 = 24J

    However, our teacher challenged us to come up with the same answer without using the mass; as it is, I've only managed to partly complete it, though I'm not sure that I'm on the right track.

    If anyone could help me determine this problem without using the mass, it would be greatly appreciated!

    I might have gone too far with my explanations, but keep in mind I'm new here. Thanks a lot!
  2. jcsd
  3. Apr 16, 2005 #2
    Without the air resistance the ball would have hit the ground at [itex] v = \sqrt{2gh} = 14m/s. [/itex]However it hit with 10m/s. The difference in velocity is attributed to the energy taken by air resistance.

    Ideal kinetic energy from gravity by:

    [tex] \Delta KE = mgh = 98m J [/tex]

    Actual kinetic energy:

    [tex] \Delta KE = mv^2/2 = 50m J [/tex]

    Energy dissipated by air resistance = 48m J

    I dont think you can go any farther than that.
  4. Apr 16, 2005 #3
    Ok, so if I correctly understand, you attributed the mass as being 1, or you simply neglected it.

    I also came up with 14m/s for the velocity, and then used that to prove that [tex]9.8 m/s^2[/tex] was indeed the regular acceleration.

    It's odd that the answer you get to is double his given answer; there must be a reason. He gives the mass of the ball as being .5kg, so, you're obviously on the right track. If we multiply the answer by .5, we get his given answer; I'm just confused as to how to get to the answer without using the .5.

    If this may further help you, my teacher told me that I had made some good process with the following work.

    With the given information, you can determine the acceleration:

    [tex]Vf^2 = Vi^2 + 2ad[/tex]
    [tex]10^2 = 0 + 2a10[/tex]
    [tex]5 m/s^2 = a[/tex]

    Now, technically speaking, the regular acceleration should be [tex]9.8 m/s^2[/tex]. He told me that, using the the ratio of these two accelerations, I should be able to follow other steps in order to come up with the answer; he left me the problem at that. Hopefully, this should help. Thanks a lot, it is greatly appreciated.
    Last edited: Apr 16, 2005
  5. Apr 16, 2005 #4


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    I must say I`m stumped as well.
    I can't think of a reason why it would NOT depend on the mass of the ball.
    Since every object has the same acceleration as a result of gravity in the absence of air resistance, every object would hit the earth with the same speed.
    The resistance force due to the air depends on the shape of the object (a ball in this case) and its velocity, but not on its mass. It requires a greater force to slow down an object with a greater mass, so naturally, the energy difference (work done by the air) must depend on the mass.
  6. Apr 16, 2005 #5
    I came up with the same reasoning as you did. I gave a little more information at the same time as you posted (see above your post), so perhaps that little bit might be of some help. Thanks!
  7. Apr 16, 2005 #6


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    No, whozum simply let the mass be undetermined and called it m. The rest is identical with your first calculations:

    [tex]E_i=mgh, \qquad [/tex]
    Pluggin in the data:[itex]\Delta E=m(9.8)(10)-\frac{1}{2}m10^2=48(m) J[/itex]. So it depends on m.

    That formula is only valid for constant acceleration.
    So either you should assume the air resistance is constant during the fall, in which case you can calculate what it would be (and find that it depends on the mass of the object anyway and receive the same answer :rolleyes: ).
    Or, there is no other way. Suppose there was a way to calculate the work done by the air, getting 24 J without using the mass of the ball. That wouldn't make much sense right? 24 J is correct, as your teacher said, now you could do the same problem with a mass of 1 kg and get 48J, or 5 kg and get 240J. If there was a way to calculate it without knowing the mass, the answer must be independent of the mass.

    You may have misunderstood the teacher? If not, I`m very curious about what he has to say.
  8. Apr 16, 2005 #7
    No I left it as a variable, notice the m's floating around. If you say m = 0.5kg then you get 24J, the answer you had. You NEED m to get an energy
    The net acceleration is not 9.8m/s^2. If it was, it wouldve hit at 14m/s. Instead the net acceleration can be found by finding the impact speed, flight time, and dividing them like I did.
    My conclusion isnt really an answer as it is an equation to find the energy of an object under those conditions. See above.

    You know what, that might work. Hang on.
  9. Apr 16, 2005 #8
    Nevermind, you can't calculate the energy without the mass regardless of the acceleration. I was thinking if you could find the GPE normally, and then multiply by the ratio of the accelerations you could find the GPE with air resistance. Subtracting the two should get the right answer, but I dont believe thats doable.
  10. Apr 16, 2005 #9
    Ok, I believe that I understand your reasoning.

    According to what he told me, I didn't even need to consider the mass when calculating the answer. Nonetheless, he did mention that the [tex]5 m/s^2[/tex] and the [tex]9.8 m/s^2[/tex] were important. He also mentioned the fact that I had to use them in accordance to ratios, so, your last statement might be correct? :confused:

    By the way, what does GPE stand for?

    I am definitely stumped, however, he seems to be positive that there's a solution. Well, at least I'm hoping that there's a solution. Thanks!
    Last edited: Apr 16, 2005
  11. Apr 16, 2005 #10
    GPE = Gravitational Potential Energy

    Energy depends on mass. I dont tihnk theres a numerical solution for this, perhaps the answer he's looking for IS [itex] 48(m) J [/itex].
  12. Apr 16, 2005 #11
    Ok, thanks; I thought it meant that, but I wasn't sure.

    Perhaps he might actually be looking for that, but I can't be sure; I won't get an answer before Monday, so, in the mean time, if anything else pops in your mind (or anyone's mind), please let me know. Thank you very much!
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