Question

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released

Answer 1

Answer:

171.5m

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

hence their separation 1.00 s after the second object is released is 171.5m