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Initial Velocity of a kicked rock

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data

    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 3.00s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

    2. Relevant equations
    Well to start off I have a graph which I don't know how to put it in here. To get things started Im currently taking General Physics and it seems a bit confusing than I thought.

    Teacher gave us this:

    r= x= ro + Vox t
    y= Yo + Voy t + (gt^2)/2


    v= Vfx= Vox
    Vfy= Voy + gt

    Im not sure if theres more to this ^

    To my understanding where r= displacement if im correct, and v is velocity(speed)

    it seems that i plug in numbers and calculate them but im having problem understanding the letters what they represent so i can't get to my solution or explain that, thats my solution. It may sound confusing

    3. The attempt at a solution

    Height= 40m
    a=g=9.8 m/s^2
    speed of sound= 342m/s
    Time=3s

    r = r0 + Vot + (gt^2)/2
    = 0 + 342 m/s *3.0s -4.9m/s^2 * 3.0s^2
    = 984.9 m

    so then I used:
    x = x0 Voxt
    984.9m = 0 + Vox * 3s
    Vox= 328.3 m/s

    and:
    y= y0 + Voyt + (gt^2)/2
    40m= 0 + Voy*3 + -4.9m/s^2 * 3.0s^2
    Voy= -28m/s

    and im stuck :( any help on what/how to get intial Velocity. Thanks in advance

    Im new to forums as well :P
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    I'm finding it difficult to follow your solution... First step... what is the distance between the place where the soccer ball fell and where the soccer player is... hint: use the speed of sound.
     
  4. Sep 14, 2007 #3
    well thats the thing how would i find the distance? oh and by the way its a rock (ball) :P
     
  5. Sep 14, 2007 #4

    learningphysics

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    lol, why is he kicking a rock? hint: they give the speed of sound... they give the time it takes to hear the sound... what distance does the sound travel?
     
    Last edited: Sep 14, 2007
  6. Sep 14, 2007 #5
    so speed of sound x time = distance?

    343m/s * 3s = 1029 m

    correct me if im wrong :(
     
  7. Sep 14, 2007 #6

    Kurdt

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    That is correct. Now how do you think you can find the horizontal distance?
     
  8. Sep 14, 2007 #7
    put it in Triangle :p

    so 40m^2 Height x Hd^2 = 1029^2

    Hd^2 = 1029^ + 1600

    Hd= 32 m

    correct?
     
  9. Sep 14, 2007 #8

    Kurdt

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    No thats not correct. Remember that the hypotenuse of a right angled triangle is given by [itex] a^2 = b^2 + c^2 [/itex]. That means you need to subtract the vertical height squared from the hypotenuse squared and square root it. Thats a horrible sentence so I'll write it in formula.

    [tex] b = \sqrt{a^2 - c^2} [/tex]

    Next in the problem you'll have to find the time of flight of the rock. How do you think you would find that?
     
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