oldhw 08 - alexander (jra2623) oldhomework 08 Turner...

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Unformatted text preview: alexander (jra2623) oldhomework 08 Turner (92510) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A soccer player kicks a rock horizontally off a 39 . 4 m high cliff into a pool of water. The acceleration of gravity is 9 . 8 m / s 2 . If the player hears the sound of the splash 3 . 79 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m / s. Correct answer: 114 . 602 m / s. Explanation: Given : y i =- 39 . 4 m , g = 9 . 8 m / s 2 , and t splash = 3 . 79 s . The rock follows a parabolic path and the sound comes back in a straight line. x y x y d The time of flight can be obtained from the vertical motion: y = v iy t- 1 2 g t 2 =- 1 2 g t 2 since v iy = 0, so t flight = radicalBigg- 2 y g = radicalBigg- 2 (- 39 . 4 m) 9 . 8 m / s 2 = 2 . 83563 s . Since the player hears the sound of the splash 3 . 79 s after the kick, the time required for the sound to travel straight back to the player is t sound = t splash- t flight = 3 . 79 s- 2 . 83563 s = 0 . 954367 s . and the straight line distance from the point the rock hits the water to the player is d = v sound t sound = (343 m / s) (0 . 954367 s) = 327 . 348 m . Since d 2 = ( x ) 2 + ( y ) 2 , the horizontal distance the rock travels is x = radicalBig d 2- ( y ) 2 = v ix t flight . Thus v ix = radicalbig d 2- ( y ) 2 t flight = radicalbig (327 . 348 m) 2- (39 . 4 m) 2 2 . 83563 s = 114 . 602 m / s ....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.

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oldhw 08 - alexander (jra2623) oldhomework 08 Turner...

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