Question

Carol drops a stone in a mine shaft 122.5 metres deep.How

Answer 1

Answer:

 T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken  = ?

time taken by the stone to reach at the bottom

using equation of motion

$s = u t + \dfrac{1}{2}gt^2$

initial speed , u = 0 m/s

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 122.5}{9.8}}$

       t = 5 s

time taken by the sound to travel

    d =v x t

 $t = \dfrac{d}{v}$

 $t = \dfrac{122.5}{340}$

    t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

 T = 5.36 s