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

Physics

1 answer:

Verizon [17]3 years ago

7 0

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

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