A soccer player kicks a rock horizontally off a 40-m-high cliff into a pool of w

Question

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

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Explanation / Answer

Since the player kicked the ball horizontally, vertical velocity of the ball is zero. Time taken to go down by 40 m with zero initial velocity = t => 40 = (1/2)gt^2 => t = v(80/9.81) = 2.86 s. => time taken by sound to reach the player = 3.00 - 2.86 s = 0.14 s and distance covered by sound = 343 * 0.14 m = 48.02 m => horizontal distance covered by the ball = v[(48.02)^2 - (40)^2] m = 26.56 m. This distance was covered by the ball moving at constant horizontal velocity, v, in the same time as it took to reach down to the surface of water, i.e., 2.86 s. => v = 26.56/2.86 = 9.289 m/s.

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