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Two dimensional motion problem: speed of sound as a factor?

  1. Sep 1, 2006 #1
    I am stumped by the following problem, I can't quite figure out what to do with the information about the speed of sound:
    A soccer player kicks a rock horizontally off a 31.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.1 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

    I don't think it is implying that the object was traveling at the speed of sound, so I'm not really sure how it figures into the problem, since it does not give a delta X value.
    The known quantities are delta Y = -31 m and t= 3.1, so using
    delta Y= Voyt + (1/2)gt^2 and substituting zero for Voy, I solved for t and got 2.525 seconds.
    Using Vy=Voy + at, I got that Vy = 24.647 m/s.

    That is where I'm stuck, any advice on where to proceed from here?

    Thank you for your time!
     
  2. jcsd
  3. Sep 2, 2006 #2

    Chi Meson

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    The rock travels for a while as a projectile, then it makes the splash, then it takes some time for the sound of the splash to get back to you.

    The total time for both parts is 3.1 s.

    It sounds harder at first until you realize that the time for the rock to fall does not depend on the initial horizontal speed.
     
  4. Sep 2, 2006 #3

    andrevdh

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    The 2.52 seconds is the elapsed time from kickoff until the stone hits the water. The remaining time is for the sound of the splash to reach the player. This sound had to travel along the line of sight, that is from the entry point in the water up to the player.
     
  5. Sep 2, 2006 #4
    Thanks for the responses, I appreciate it. So if there is a 0.575 second lag time between when the object hits and when the splash is heard, how do I use that to figure out the initial velocity? I'm sorry if I'm being dense with this...I've got the list of kinematic equations in front of me but can't figure out what my next step should be to solve for Vox and then Vo. I tried using tan theta = Vy/Vx, using 180 degrees since it was kicked along a horizontal, but tan theta is zero...
    Any tips are extremely welcome and appreciated!
     
  6. Sep 2, 2006 #5

    Chi Meson

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    You have speed (of sound) and time. What must be the net displacement?

    What's the vertical displacement?

    Take it from there..
     
  7. Sep 2, 2006 #6
    Net displacement should be speed times time, but I am trying to solve for the speed of the object, and what I have is the speed of sound. The vertical displacement is -31 meters, so if it hit after 2.525 seconds then it dropped with an average velocity of 12.277 m/s, but that is almost exactly half of what I had calculated Vy as being using Vy=Voy + at. I'm sorry that I'm having such a hard time "getting" this, but without an angle or any horizontal values, I feel like I keep hitting a wall.
     
  8. Sep 2, 2006 #7

    andrevdh

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    You can calculate the line of sight distance along which the sound travels back to the player with the remaining time and the speed of sound. This distance forms a right angled triangle with the height and the total x-displacement of the rock. See if you can calculate this x-displacement.
     
  9. Sep 3, 2006 #8
    I tried to get the line of sight distance the way you suggested by using 343 m/s times 2.525 s, to cancel out meters, and got 866.075 m. I used the pythagorean theorem with the delta Y value to get delta X, 31m^2 + delta X^2 = 866.075^2. I came up with an answer of 865.52 m for the delta X, but that seems much too big. I tried that answer on Webassign anyway, and it told me I was off by an order of magnitude. Just out of curiosity I moved the decimal place over, using 86.552, and it told me that now I am within 10% of the correct answer. Apparently this happened by dumb luck, because I still don't know how to proceed. I am sure I am making this much harder than it should be, again I apologize, I really appreciate your patience, I am not trying to ask anyone to give me an answer but am just really hitting a wall about the appropriate steps to take, it seems like everything I try just brings me to a new dead end.
    Thank you, again any tips are extremely appreciated.
     
  10. Sep 3, 2006 #9
    Remeber that the time taken by sound to reach you is 0.575 secs and not 2.525 secs . Multiplying with the speed of sound, you can get the net displacement of the projectile (hypotenuse). Use Pythagoras' theorem to find Xy .
     
  11. Sep 3, 2006 #10
    I got it! Thanks, I think my main problem was using which t value (total time, falling time, or the difference) to use where. I multiplied the difference times the speed of sound, like you said, and used that as my delta X value, then used delta X = Voxt to solve for the initial horizontal velocity, using the falling time t value, and finally got the correct answer.
    Thanks so much to everyone for your time and patience, it means a lot!
     
  12. Sep 3, 2006 #11
    what was the answer you ended up getting for this problem??
     
  13. Sep 3, 2006 #12
    The question asked for initial speed, and I got 79.78 m/s.
     
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