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MC363A

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- Thread starter MC363A
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- #1

MC363A

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- #2

stunner5000pt

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anyway here's how i would solve it

for the speed of sound to travel back up the distance traveled was teh height of the cliff so [tex] d = 340 t_{sound} [/tex]

now for the fall

v1 = 0

v2 = ?

a = -9.8m/s^2 (gravity points down, i took down to be negative, it doesn't matter whichever way you take it )

[itex] t_{fall} = ? [/itex]

now relate what u have to what you don't have. Also what u don't need can be omitted in that equation you form. How are [tex] t_{fall} and t_{sound} [/tex] related? Tfall is the time it took for hte stone to fall into the water. Tsound is the time it took for the sound to reach you. How long was this entire trip?

- #3

MC363A

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- #4

stunner5000pt

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[tex] d = v_{1} t + \frac{1}{2} at_{fall}^2 [/tex]

v1 = 0 so

[tex] d = \frac{1}{2} (-9.8) t_{fall}^2 [/tex]...1

the speed of sound part

[tex] d = v_{sound} t_{sound} [/tex]

[tex] d = 340 t_{sound} [/tex]...2

so

[tex] t_{sound} + t_{fall} = 3.4 [/tex]...3

combine 1,2, and 3 and should should be able to solve for one of the times. Thus u can solve for the distance.

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