question_answer

A stone is dropped into a well 44.1 m deep. The sound of the splash is heard 3.13s after the stone is dropped. Find the velocity of sound in air (\[g=9.8m{{s}^{-2}}\])

Answer:

Let us first calculate time taken by stone to fall through 44.1 m. \[u=0\] \[s=44.1m\] \[a=g=9.8m{{s}^{2}}\] \[t=?\] Using \[s=ut+\frac{1}{2}a{{t}^{2}},\] We have \[44.1=0+\frac{1}{2}\times 9.8\,{{t}^{2}}\] \[=4.9\,{{t}^{2}}\] \[\Rightarrow \]\[{{t}^{2}}=9\] or \[t=3s\] The remaining time i.e.\[3.133=0.13s\]is taken by sound to go up through 44.1 m. \[\therefore \] \[t=0.13s\] \[s=44.1m\] \[v=\frac{s}{t}\] \[=\frac{44.1\,m}{0.13\,s}=\frac{4410}{13}=339.23\,m{{s}^{-1}}\]