Question

You drop a stone into a deep well and hear it hit the bottom 1.70 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about 343 m/s in air. How deep is the well?

Answer 1

please help with the steps..pleaseee..thank you

Answer 2

t= t1 + t2

\[t1 = \sqrt{2h \div g}\]

t2 =h/v

Answer 3

Sorry, I got held up quite a bit with another question.

First note the following kinematic equation:
\[h=\frac{1}{2}gt^2\]Solve that equation for t_1 to get:
\[t_1=\sqrt{\frac{2h}{g}}\]That's basically how long it takes an object to hit the bottom of the well.

Now we need to know the formula for the time for the sound to reach the top of the well. That will simply be:
\[t_2=\frac{h}{V_{sound}}\]
The total time will obviously be the time it takes to go down + the time it takes to come back up so:
\[t_{final}=t_1+t_2\]
Plug in what you know to get this equation:

\[1.70s=\sqrt{\frac{2h}{9.8m/s^2}}+\frac{h}{343m/s}\]
It will take a bit of algebra but all you have to do is solve for h :)

Answer 4

You should end up with an answer of ~13.5m.

Answer 5

thank you:)))))))

Answer 6

np :)