Question

A robber dropped a bag of loot into a 30 m well. If the speed of sound is 340 m/s how long after dropping the bag does he hear the splash?

Answer 1

First we have to determine the time for the bag fall by the kinematic equation,

`s=ut+(1)/(2) g t^(2)`

Here, u is initial velocity which is zero, g is acceleration due to gravity its value is 9.8 m/s2 and s is displacement and its value is given 30 m.

Therefore,

`30 \  m =0* t + (1)/(2) (9.8 m/s^2) t^2 \\\\ t^2= (30)/(4.9) \\\\ t= √(6.1 s) = 2.46 \ s`

As sound travels at 340 m/s, so the time for the sound to travel back up the well,

`T = (distance )/(speed) = (30 m)/( 340 m/s) =0.0882 s`

Thus , the splash is heard at
`t+T= 2.46 + 0.0882 = 2.55 \ s`.

Therefore, splash is heard about 2.55 s after the bag is dropped.