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Speed of sound, easy question

  1. Jul 16, 2013 #1
    Hello.

    1. The problem statement, all variables and given/known data

    A person holds a rifle horizontally and fires at a target. The bullet has a muzzle speed of 20 m/s and the person hears the bullet strike the target 1.00s after firing it. The air temperature is 72 degrees F. What is the distance to the target?


    2. Relevant equations

    speed of sound = 331 + 0.6*T (where T =temp in celcius)


    Note

    I tried solving the problem, and got a different answer than from my solutions manuel. I was just wondering how everyone else would solve it.
     
  2. jcsd
  3. Jul 16, 2013 #2
    I think you should use the more general formula for the speed of sound that is [itex]sqrt(gamma*RT/M)[/itex]. Moreover, show your attempt with the formula you used.
     
  4. Jul 16, 2013 #3

    haruspex

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    ... and the answer from the solutions manual.
    (Fwiw, I would have thought drag on the bullet would be more significant than variations in the speed of sound.)
     
  5. Jul 16, 2013 #4
    (°F - 32) * 5/9 = °C
    (72 - 32) * 5/9 = 22.22 °C
    22.22*0.6 + 331 = 344.33 m/s

    it takes one second for the bullet to reach the target and then for the sound to travel back to you so using
    d=vt
    t(1s) = (d/344.3)+(d/20)
    factor out a d

    1s = d((1/344.3)+(1/20))
    d = 1/((1/344.3)+(1/20))
    d = 18.9 m

    Hope I did that correctly and hope you could follow along! This should be the answer you get
     
    Last edited: Jul 16, 2013
  6. Jul 16, 2013 #5
    Not sure how to delete a post so i guess this will stay here emarrasingly indefinitely
     
    Last edited: Jul 16, 2013
  7. Jul 16, 2013 #6

    haruspex

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    The main problem with your post is that it is not how this forum works. We don't do people's homework for them. The idea is to nudge in the right direction and point out flaws in arithmetic, algebra or reasoning. So far, oneplusone has not posted enough detail to comment on (beyond that comment).
     
  8. Jul 16, 2013 #7
    ok sorry :/
     
  9. Jul 16, 2013 #8
    Hello,

    Sorry, i wasn't able to get onto a computer until now.
    I did what cmcraes did, except I subtracted instead of added.
    Is the reason why you added that the bullet pushes the air which creates the sound, so the sound is the combined velocity of the bullet and the sound? or why do you add?
    And this isn't a homework question, (I'm trying to learn physics on my own).
     
  10. Jul 16, 2013 #9
    You add because you have to remember that From the time you pull the trigger, to when the sound reaches your ears is one second. So you add the time it takes the bullet to hit the target to the time it takes the sound to reach you from the target.
    (since that equals 1, and t=d/v, than (d/V_sound + d/V_Bullet) must equal one!) hope that made sense! :)
     
    Last edited: Jul 16, 2013
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