Question

A Steel ball dropped into a well 44.1 m deep the sound of splash is heated 3.13sec after the Steel ball dropped. find the velocity of sound in air​

Answer 1

As ball splashed after covering a distance of 44.1 m at 3.13 s

=> S= ut + 1/2 gt^2

=> 44.1 = 0 + 1/2 x 9.8 x t^2

=> 44.1/4.9 = t^2

=> 9 = t^2

$=> t = \sqrt{9}$

$\bold{ => t = 3 \: s}$

Hence, the ball reached there after 3 s.

So,

=> Time = Final time - Initial time

=> T = (3.13 - 3) s

=> T = 0.13 s

As we know that,

=> Velocity = Distance/Time

=> V = 44.1/0.13

$\bold{=> V = 339.23 m{s}^{-1}}$