A woman is swimming when she hears the underwater sound wave from an exploding ship across the harbor. She immediately lifts her head out of the water. The sound wave from the explosion propagating through the air reaches her 4.00 s later. How far away is the ship? Assume that the water temperature is 25°C and the air temperature is 23°C.
you will have to use your equations to find velocity of sound in water Vw and air Va at those temperatures
d = Va t so t = d/va
d = Vw (t-4)
d = Vw (d/Va - 4)
4 Vw = d (Vw/Va -1)
d = 4 Vw /(Vw/Va-1)
d = 4 /(1/Va -1/Vw)
Okay, velocity of air is 345.03 m/s and velocity of water is 1500m/s
I did
D=Va(t) D=345.03m/s(4.0s) and got 1380.12m
Then:
1380.12m=1500m/s (t-4) and I got t=4.92s
But then I don't know what to do after that
but all it asked for was d, 1380 meters
d = 4 /(1/Va -1/Vw)
d = 4 /(.00289 - .000667)
= 4/.0022
= 1818 m