Question

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.

Answer 1

is t total time?

Answer 2

the sound is returned exactly 3 seconds after the stone is dropped. And I believe the answer should be less than 550 ft. deep.

Answer 3

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)

Answer 4

(and for the record you don't need calculus)

Answer 5

\[h=\frac{1}{2}g*t _{1}^{2}\]
\[h=v*t _{2}\]
\[t _{1}+t_{2}=3\]

Answer 6

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...

Answer 7

\[\frac{1}{2}g*t _{1}^{2}=v*t _{2}\]

Answer 8

\[\frac{1}{2}g*t _{1}^{2}=v*(3-t _{1})\]

Answer 9

@cinar dont mess with more unknown, is simple just take 2 unknown...
equation is easy..but calculation really mind screwing..

Answer 10

\[t _{1}=2.89\]
h=16*(2.89)^2=132.9 ft

Answer 11

everything is known, v=1126 g=32 just find t