Ralated rates

• Nov 17th 2007, 07:45 PM
btwest69
Ralated rates
A stone is dropped into a deep, dark mine. A clunk is heard 7 seconds later. Estimate the depth of the shaft in feet. Ignore air resistance. Take the speed of sound as 1000 feet/second.

Velocity of sound= 1000 ft/sec
d = 16t^2

can anyone help me to solve this question?
Thank you.
• Nov 17th 2007, 08:01 PM
Jhevon
Quote:

Originally Posted by btwest69
A stone is dropped into a deep, dark mine. A clunk is heard 7 seconds later. Estimate the depth of the shaft in feet. Ignore air resistance. Take the speed of sound as 1000 feet/second.

Velocity of sound= 1000 ft/sec
d = 16t^2

can anyone help me to solve this question?
Thank you.

see here

it is a very similar problem, do you understand it?
• Nov 17th 2007, 08:41 PM
btwest69
Thank you
I get it now, i just need to solve for s.

$
\boxed{ \frac{\sqrt{s}}{4}+ \frac{s}{1000} = 7 }
$
• Nov 17th 2007, 08:44 PM
Jhevon
Quote:

Originally Posted by btwest69
Thank you
I get it now, i just need to solve for s.

$
\boxed{ \frac{\sqrt{s}}{4}+ \frac{s}{1000} = 7 }
$

yes.

i hope you really understood what happened and didn't just plug in the relevant numbers into TPH's equation.

could you, for instance, tell me how he came up with the $\frac {\sqrt{s}}4$?
• Nov 17th 2007, 08:59 PM
btwest69
s = 16 t^2

t= (s/16)^1/2

$t=\frac {\sqrt{s}}4$
• Nov 17th 2007, 09:06 PM
Jhevon
Quote:

Originally Posted by btwest69
s = 16 t^2

t= (s/16)^1/2

$t=\frac {\sqrt{s}}4$

good enough for me.

there is an issue with signs though, but it depends on how you think about the problem. but you're good. good job