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 (\[\text{g}=\text{9}.\text{8m}{{\text{s}}^{-\text{2}}}\])

Answer:

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