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Projectile Motion

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A soccer player kicks a rock horizontally off a 40 m high cliff into a pool of water. If the player hears the sound of the splash 3 seconds later, what was the intial speed given to the rock? Assume that the speed of sound in air is 343 m/s


    2. Relevant equations
    V_f = V_i - gt
    X_f = x_i + vt - 1/2gt^2 (which becomes zero)
    Y_f = Y_i + vt (which becomes zero) -1/2gt^2
    y direction velocity = v_i sin*
    x directino velocity = v_i cos*

    3. The attempt at a solution

    Well first I tried using -1/2gt^2 to find the y_f, just to check its validity... and I got y_f = -44.1 m, which is lower than the cliff...
    And I also found t = 2.857 to get to y_f = -40

    But I'm so confused with so little data. I know that:
    y_i = 0 m
    y_f = -40 m or -44.1 m
    t = 3s
    x_i = 0m
    x_f = ?
    a_x = 0 m/s^2
    a_y = -9.8 m/s^2
    v_x = ? m/s
    v_y = 0 m/s

    Could someone possibly guide me a little bit, since I'm not understanding the equations properly, I guess. Thanks ^_^
     
  2. jcsd
  3. Sep 19, 2007 #2

    learningphysics

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    How long does it take to drop the 40m?
     
  4. Sep 19, 2007 #3
    2.857 seconds
     
  5. Sep 19, 2007 #4

    learningphysics

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    exactly. so for how long does sound travel... and what distance does it travel?
     
  6. Sep 19, 2007 #5
    sound travels at a rate of 343 m/s...
    v = d/t
    d= vt
    So it travels (353 m/s)(2.857 s) = 1008.5 m?
     
  7. Sep 19, 2007 #6

    learningphysics

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    yes, but the sound only begins its travel the moment the stone hits the water... that is at t = 2.857
     
  8. Sep 19, 2007 #7
    So if the soccer player hears the sound 3s after he threw the stone, that must mean the sound traveled in that .143 second interval
    Then that would mean it travels at (343 m/s)(.143 m) = 49.05 m?
     
  9. Sep 19, 2007 #8

    learningphysics

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    Yes... now can you get the horizontal distance between the cliff and where the stone fell in the water?
     
  10. Sep 19, 2007 #9
    Um... I'm not sure how to do that part. I know an equation
    x_f = x_i + (v_i)(t) - .5gt^2 = (v_i)(t).. which has two missing variables
    t would be 2.857 s because the stone reaches the water at that point...
    I don't think 49.05 fits in that equation either..
     
  11. Sep 19, 2007 #10
    well, actually.. if I use 49.05 m minus the distance the stone traveled between 2.857 s and 3 s, I can find the horizontal distance between the cliff and the point where the stone reached the water. But without a velocity variable I'm lost..
     
  12. Sep 19, 2007 #11

    learningphysics

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    Use the pythagorean theorem... the 40m vertical and the horizontal displacement, and the 49.05m form a right triangle.

    the sound travels in a straight line back to the original point.
     
  13. Sep 19, 2007 #12
    oh! That makes sense now. I know how to complete the problem. thank you for your help very very much. I did not realize the sound traveled straight back such as the hypotenuse of a triangle.
     
  14. Sep 19, 2007 #13

    learningphysics

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    no prob.
     
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