Posted by **Anonymous** on Sunday, October 11, 2009 at 12:06am.

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.

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**MathMate**, Sunday, October 11, 2009 at 12:27am
Cliff Height, H = -42.9 m

Time to drop down cliff, t

H=0*t + (1/2)*(-g)t²

t=√(2*H/(-g))

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.

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**Anonymous**, Sunday, October 11, 2009 at 1:22am
I got the right answer.

However, the equation to find the horizontal distance would be:

h = square root of (D^2 - H^2)

- physics -
**MathMate**, Sunday, October 11, 2009 at 7:52am
Yes, you are correct, since the oblique distance is the hypotenuse.

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