P18_007

# P18_007 - t/v s d(1/v 2 s d 2 Now multiply by gv 2 s and...

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7. Let t f be the time for the stone to fall to the water and t s be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hearing the splash is t = t f + t s .I f d is the depth of the well, then the kinematics of free fall gives d = 1 2 gt 2 f ,or t f = p 2 d/g . The sound travels at a constant speed v s ,so d = v s t s ,or t s = d/v s .Thu sthe total time is t = p 2 d/g + d/v s . This equation is to be solved for d .Rew r i tei ta s p 2 d/g = t d/v s and
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Unformatted text preview: t/v s ) d + (1 /v 2 s ) d 2 . Now multiply by gv 2 s and rearrange to get gd 2 − 2 v s ( gt + v s ) d + gv 2 s t 2 = 0. This is a quadratic equation for d . Its solutions are d = 2 v s ( gt + v s ) ± p 4 v 2 s ( gt + v s ) 2 − 4 g 2 v 2 s t 2 2 g . The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d = 40 . 7 m is obtained....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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