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anonymous

  • 5 years ago

Anyone have a simple solution to this one using calculus? A stone is dropped into a shaft and the return echo is timed at exactly 3 seconds. We know the speed of sound is 1126 ft/sec and gravitation acceleration is 32 ft/sec/sec. What is the depth of the shaft? Thanks for your help.

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  1. anonymous
    • 5 years ago
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    is t total time?

  2. anonymous
    • 5 years ago
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    the sound is returned exactly 3 seconds after the stone is dropped. And I believe the answer should be less than 550 ft. deep.

  3. mathmate
    • 5 years ago
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    Let H=depth of shaft (feet) t =time for stone to descend (in seconds) assuming no air resistance, g = acceleration due to gravity (32 ft/s^2) v= velocity of sound (ft/s) v0=initial velocity of stone (0 ft/s) T = time for echo to go up to ground By kinematics equations H=v0*t+(1/2)gt^2 Solve for t t=sqrt(2H/g) By speed of sound, time required for echo T=H/1126 Since t+T=3 seconds, we have sqrt(2H/g)+H/1126=3 Solve for H to get 132.9 ft. (check my arithmetic)

  4. JamesJ
    • 5 years ago
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    (and for the record you don't need calculus)

  5. anonymous
    • 5 years ago
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    \[h=\frac{1}{2}g*t _{1}^{2}\] \[h=v*t _{2}\] \[t _{1}+t_{2}=3\]

  6. anonymous
    • 5 years ago
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    let dipth is s.. so s= 1/2 g t^2.. and s/1126 = 3 - t now you have 2 equation and two unknown solve it...

  7. anonymous
    • 5 years ago
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    \[\frac{1}{2}g*t _{1}^{2}=v*t _{2}\]

  8. anonymous
    • 5 years ago
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    \[\frac{1}{2}g*t _{1}^{2}=v*(3-t _{1})\]

  9. anonymous
    • 5 years ago
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    @cinar dont mess with more unknown, is simple just take 2 unknown... equation is easy..but calculation really mind screwing..

  10. anonymous
    • 5 years ago
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    \[t _{1}=2.89\] h=16*(2.89)^2=132.9 ft

  11. anonymous
    • 5 years ago
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    everything is known, v=1126 g=32 just find t

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