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Finding distance traveled of a dropped object

  1. Dec 16, 2007 #1
    [SOLVED] Finding distance traveled of a dropped object

    1. The problem statement, all variables and given/known data
    Imagine that someone dropped a firecracker off the roof of a building and heard it explode exactly 10 s later. Ignoring air friction, taking g = 9.81 m/s2 and using 330 m/s as the speed of sound, calculate how far the cherry bomb had fallen at the very moment it blew up.

    2. Relevant equations
    Kinematic equations, Straight-line motion equations

    3. The attempt at a solution
    First I found total distance traveled by the firecracker using the 2nd kinematics where vi=0. The total distance found after the firecracker traveled for 10s was 490.5m. Since the speed of sound travels at 330m/s, I did 490.5m/(330m/s) to find the time it takes sound to travel 490.5m. The time found was 1.48636s. I subtracted this time from the total time which was 10s. This gave the time the firecracker traveled at the moment it blew up which was 8.51364s. I plugged in this time into the 2nd kinematics again where vi=om/s and found a distance of 355.288m but it was incorrect......any idea on where I went wrong?
  2. jcsd
  3. Dec 16, 2007 #2
    You assumed the the firecracker fell for the entire 10 seconds before it exploded. 10s = the time it fell+the time to reach the observer after it exploded.
  4. Dec 16, 2007 #3
    Huh? I thought I accounted for the time it took to reach the observer. Since the observer heard it 1.48636s after the moment of explosion. I subtracted 10s which was total time by the time it took to hear the explosion which I thought would give the time traveled from rest to moment of explosion.
  5. Dec 16, 2007 #4
    You found that it takes 1.48636s for a sound wave to go 490.5 meters. You then subtracted that from the 10 seconds and got 8.51364s. You can't do this because you all ready made the assumption that it took 10s before the firecracker explodes to reach 490.5m, you then said that the firecracker fell for 8.51364s before exploding which contradicts your first assumption.
  6. Dec 16, 2007 #5
    Would this mean the time I found isn't the actual time it takes to reach the observer?
  7. Dec 16, 2007 #6
    Yes. You would go about solving the problem by seeing that the total displacement (not really displacement but we can consider it to be so) is 0 over the entire 10 seconds.

    [tex]0 = \frac{1}{2}gt_{1}^2 - 330t_{2}[/tex]


    [tex]10s = t_{1}+t_{2}[/tex]
    Last edited: Dec 16, 2007
  8. Dec 16, 2007 #7
    The total distance of 490.5m is right though correct? I just want to know which part of my work is correct so I can work with it.
  9. Dec 16, 2007 #8
    That's not right either because it takes the entire 10s to fall that far, which would mean you would need more time to have the sound reach the observer, which would make the total time > 10
  10. Dec 16, 2007 #9
    You need to add the time it traveled and the time the sound took to go back to the observer.
  11. Dec 16, 2007 #10
    To find the time it takes to travel to observer wouldn't I need to find the distance it traveled after 10s? Which I thought would be 1/2(9.8m/s^2)(10s)^2 which gave me the 490.5m.
  12. Dec 16, 2007 #11
    No because you heard it explode exactly 10s later, which means it needed to travel a distance explode, then the sound had to reach the observer all in 10 seconds which means that it had to explode before t=10 since if it did explode at t = 10 it would talk more then 10 seconds for the sound to reach the observer.
  13. Dec 16, 2007 #12
    Ok, so if sound travels at 330m/s and the observer heard the firecracker at 10s. It would take sound 10s to travel X distance. If I do (330m/s) x 10s this would mean the sound of firecracker traveled 3300m in 10 seconds right, which would mean this is the distance traveled by the firecracker from drop to hearing the explosion correct? Why does it seem like these numbers don't sound right....

    Edit: Sorry didn't see what you wrote above. Going to reedit after.
    Last edited: Dec 16, 2007
  14. Dec 16, 2007 #13
    10s is the total time. This means that part of the 10 seconds the object is falling, the other portion of the 10 seconds is the sound wave traveling through the air at a speed of 330m/s to reach the person who dropped the firecracker.

    No, the sound of the firecracker would travel 3300m in 10 seconds BUT that's the entire interval the problem talks about. It has to fall for some of that time too, so the firecracker has to be a distance of 0meters < x meters < 3300 meters.
  15. Dec 16, 2007 #14
    Would you be able to use simultaneous equations by solving for t2 in the second equation and plug in 10s - t1 for t2 in the first equation? Also, where did the 300t2 come from?? Not sure how you got that.
  16. Dec 16, 2007 #15
    Sorry it should be 330t2 which is the distance traveled by the sound wave in t_2 seconds.
  17. Dec 16, 2007 #16
    ok, but it is possible to use simultaneous equation correct?

    THANKS A LOT!!!! I got it using simultaneous equations.
    Last edited: Dec 16, 2007
  18. Dec 16, 2007 #17
    Yeah simultaneous equations was what I was thinking.
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