# WAVES, SPEED OF SOUND.

1. May 30, 2015

### samisoccer9

Problem: A swimmer sees a parachutist hit the water and hears the impact twice, once through the water and the second time through the air, 1.0s later. How far from the swimmer did the impact occur. Vs of air is 340m/s. Vs of water is 1400m/s and the answer in the book is 450m.

2. I know this is a simple rate question, except I cannot get it! VT = D

3. I attempted to do VT = D and just sub in the velocity values for each, but it just gives you the same answer as the velocity of course.
ex of what i tried to do: 340m/s x 1s = D, D = 340m, doesnt work.

Last edited by a moderator: May 30, 2015
2. May 30, 2015

### RUber

T isn't 1sec. It is T_air =T_water+ 1 sec.

3. May 30, 2015

### samisoccer9

How would you figure that out though?

4. May 30, 2015

### RUber

It takes a certain amount of time to get to him through the water, lets call that T. Then he hears it one second later through the air...T+1.
Now you can use the fact that
$T_{air}V_{air} =D =T_{water}V_{water}$
to solve for T, and then just put it back into the equation to find D.

5. May 30, 2015

### samisoccer9

Sorry for troubling you but could you please solve it with numbers in a reply? I attempted to use the equation you just showed me but I am still stuck/unsuccessful. It would be greatly appreciated if you could, thanks.

6. May 30, 2015

### RUber

$T_{air}=T_{water}+1$
Call $T_{water}=T$
Then based only on the information you provided...
340(T+1) = D = 1400(T).
You don't need D right away, so you have to use algebra to solve for T in the equation:
340(T+1)= 1400(T).

Once you have T, multiply by 1400 to get D.