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A soccer player kicks a rock horizontally off a cliff 42.9 m high into a pool of water. If the player hears the sound of the splash 3.16 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

  • physics - ,

    Cliff Height, H = -42.9 m
    Time to drop down cliff, t
    H=0*t + (1/2)*(-g)t²

    Time for sound to travel, ts
    = 3.16-t
    Oblique distance,D = ts s. * 343 m/s
    Horizontal distance h= √(H²+D²)

    initial horizontal velocity, u
    = horizontal distance / time to reach bottom
    = h/t

    Please check my thinking.

  • physics - ,

    I got the right answer.
    However, the equation to find the horizontal distance would be:
    h = square root of (D^2 - H^2)

  • physics - ,

    Yes, you are correct, since the oblique distance is the hypotenuse.

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