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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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.

Mathematics
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is t total time?
the sound is returned exactly 3 seconds after the stone is dropped. And I believe the answer should be less than 550 ft. deep.
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)

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Other answers:

(and for the record you don't need calculus)
\[h=\frac{1}{2}g*t _{1}^{2}\] \[h=v*t _{2}\] \[t _{1}+t_{2}=3\]
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...
\[\frac{1}{2}g*t _{1}^{2}=v*t _{2}\]
\[\frac{1}{2}g*t _{1}^{2}=v*(3-t _{1})\]
@cinar dont mess with more unknown, is simple just take 2 unknown... equation is easy..but calculation really mind screwing..
\[t _{1}=2.89\] h=16*(2.89)^2=132.9 ft
everything is known, v=1126 g=32 just find t

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