Question

If a splash is heard 4.23 second after a stone is dropped into a well 78.4 m deep then calculate the speed of sound in air.

Answer 1

For downward motion of the stone, 

Initial velocity u = 0 

Acceleration a = 9.8 m/s² 

Distance S = 74.8 m. 

Time t = ? 

S = ut + \(\frac{1}{2}\) at² 

78.4 = 0 + \(\frac{1}{2}\) × 9.8 t²

= 4.9 t² 

∴ t² = \(\frac{78.4}{4.9}\)= 16 

∴ t = 4 s 

Let ‘t’ be the time taken by the splash of sound to reach the top of the well. 

Then, 

t + t’ = 4 +t’ = 4.23 s 

∴ t = 4.23 – 4 = 0.23 s 

∴ Speed of sound in air

\(\frac{distance}{Time}\)

= \(\frac{78.4}{0.23}\)

= 340.87 m/s