First we have to determine the time for the bag fall by the kinematic equation,
[tex]s=ut+\frac{1}{2} g t^{2}[/tex]
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,
[tex]30 \ m =0\times t + \frac{1}{2} (9.8 m/s^2) t^2 \\\\ t^2= \frac{30}{4.9} \\\\ t= \sqrt{6.1 s} = 2.46 \ s[/tex]
As sound travels at 340 m/s, so the time for the sound to travel back up the well,
[tex]T = \frac{distance }{speed} = \frac{30 m}{ 340 m/s} =0.0882 s[/tex]
Thus , the splash is heard at [tex]t+T= 2.46 + 0.0882 = 2.55 \ s[/tex].
Therefore, splash is heard about 2.55 s after the bag is dropped.
