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Homework Help: Kicking rock over cliff question

  1. Oct 9, 2006 #1
    *4. 23) A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

    What I did:

    Again I drew a diagram. I let my y=-40m from where my origin is. First I calculated the sound coming back to the soccer player, as that information was given in more detail. So I know the formula X=(Vi)(t) .
    So (343)(3.00) = 1024m. So I know my hypoteneuse and my opposite side which is -40. So if I apply the pith. theorm, I get:

    1024^2 - (-40)^2 = 1023.2184 m

    So now I know my displacement.

    At this point I want the total time from the kick to when it falls into the water so that I can calculate the initial velocity. I assume that I can use this formula:

    xf = vyi t -1/2 gt^2

    I assume that vyi is 0 because the soccer player kicks it straight and than it goes down (don't know if this is a valid assumption). Also, I subtract t-3 since we want to see when the rock hits the water, not includng the sound back to the the soccer player.

    -40 = 0 - 1/2 (9.8)(t-3)^2

    -40 = -4.9 (t^2-6t+9)

    -40 = -4.9t^2 = 29.4t -44.1

    = 4.9t^2 - 29.4t = 4.1

    since I have a quadratic equation, I use a quadratic formula which gives me a time of 5.8 seconds. which is essentially time of flight. So when when I find the intial velocity, i plug it into this equation:

    x = (v)(t)

    1024 = (v) (5.8)

    = 176 m/s

    ( I don't know, but I really doubt this answer for some reason, so if someone can correct it for me, it would be much appreciated!)
  2. jcsd
  3. Oct 9, 2006 #2


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    3 seconds is not the time sound travels after the splash. It is the time it takes for the rock to hit the water after being kicked plus the time for the sound to travel back to the player. The sound does not travel anywhere near the 1km+ you have calculated. Figure out how long it takes the rock to hit the water and then use the difference between that and 3 seconds to get the sound distance.
  4. Oct 10, 2006 #3
    hey Older Dan,

    Thanks again for your help. I was just wondering something. I was speaking to a classmate this morning and he said that I can assume that the sound is negligible and that I can assume that it takes 3 seconds to hit the water. He said that the extra info about how fast the sound travelled was just extra info to confuse me. Could he be right?
  5. Oct 10, 2006 #4


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    The sound has to travel at least 40 m. At a speed of 343 m/s, this means the time it takes for the sound to travel back to the kicker is at least 0.117 sec. Does that seem negligible to you?
  6. Oct 10, 2006 #5
    oh, I guess you're right. Thanks for the correction. How did you find out that it took that long to come back?
  7. Oct 10, 2006 #6


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    jtbell found the minimum time for the sound to come back, not the actual time. I think you were on the right track at the beginning. You just failed to figure out how long it takes the rock to hit the water. Think about that. Does it make any difference how hard the rock is kicked?
  8. Oct 10, 2006 #7
    thanks Older Dan! Now that I know I'm going in the right direction, I'll try to figure out where I went wrong and post the answer sometime tommorrow. Thanks again!
  9. Oct 12, 2006 #8
    alrighty, for this problem, I figured out how Jtbell, founf the minimum value for the sound coming from the splash. so I subtracted that value from t = 3.00s and cam up with 2.88. Now, because the ball starts off in a horizontal motion on top of the cliff, I assume that my angle of orientation is 0 degrees. The only thing is that I don't understand how to get any values for my x-components.

    Can anyone help?


  10. Oct 12, 2006 #9


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    Don't use the minimum value. If you do, the initial velocity will be zero. Calculate how long it takes the rock to reach the water; this is independent of the intial velocity. Then do what you did the first time to find how far the sound has to travel, but use what is left of the 3 seconds instead of 3 seconds. That part of what you did was correct.
  11. Oct 12, 2006 #10
    i'm just not sure what I'm suppose to use to calculate for when the rock hits the water.
  12. Oct 12, 2006 #11


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    This problem is like every other projectile problem. The vertical motion is independent of the horizontal motion. The problem tells you the initial vertical velocity. You know the acceleration and the height of the cliff. You can calculate the time of flight.
  13. Oct 12, 2006 #12
    from what I understand from what you're saying. I can use this equation?:

    -40= vi(sin)(0)t-1/2(9.8)t^2 =

    -40= -4.9 t^2

    -40/-4.9 = 4^2

    t = 2.85714 sec ?

  14. Oct 12, 2006 #13


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    Except for the typo. Yes. Now use the remaining time to see how far the sound travelled and you can follow your original idea.
  15. Oct 12, 2006 #14
    ok lets see.

    So I take 2.85714 and subtract it from 3 to give me 1.4286.

    Now to find my displacement in x I use X=(v)(t):

    (343)(1.4286) = 49.00

    I can use teh kinematic equation to find xf for the y-component:

    -40 = vyi - 4.9 (1.4286)

    -40/-7.000 =

    vyi = 5.71428 m/s

    than I used the x-component equaton x=v(t)

    49.0 = vxi (1.4286)

    49.0/1.4286 =

    vxi = 34.2993

    when I take both velocities and use pith theorum, I get the initial velocity to be 34.772 m/s

    Hope this is right. Let me know :)
  16. Oct 12, 2006 #15


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    49 is not the x displacement. It is the distance from the splash point to the edge of the cliff. Go back to your first post and look at what you did. You were on the right track, but using the wrong time.

    You also need to check that time calculation. It is way off.
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