An Ultrasound equipment installed at a height of 80 m above sea water surface sends out ultrasound that returns from the Sea bed and is detected after 5.4 second. If the speed of ultrasound through sea water is 1500 M per second and that through a year is 320 M per second. Then what is the depth of the sea is that place.
Answers
Hello buddy!!
First of all this question can be solved if we consider normal incidence of the ultrasonic waves instead of oblique incidence because incase of oblique incidence we would require the angle made by ultrasonic waves with the normal at the interface
Time taken from transmission to reception is given as 5.4 seconds (total time)
Therefore,
5.4= time required to travel in water +time required to travel in air
We know time = Distance /Speed
And the distance above sea level is given to be 80 m
Therefore by data,
5.4= (80*2)/320+(d*2)/1500
Where d is the depth of water
2.7=80/320+d/1500
d=(2.7-80/320)*1500
d=2.45*1500
d=3675
Hence the depth of water is 3675 meters
Hope this helps!! ^_^
The total time taken to detect = 5.4 S. We know, time = distance / speed.
Hence putting the given value we get the time = 160/ 320 + 2h/ 1500
Therefore, 5.4 = ½ + h/ 750 Hence, h = 367.5 meter. Here h is the depth of the sea that is calculated by the echo ranging method.