How far did the parachutist fall before hitting the water?

[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 = D3. 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, doesn't work.

 

[RUber]

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

 

[samisoccer9]

How would you figure that out though?

 

[RUber]

It takes a certain amount of time to get to him through the water, let's 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.

 

[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.

 

[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.