Question

An Explosion takes place at a distance of 3200 m away from a sound detector that detects only the sound coming from the ground.It detects two signals at 0.64 sec and 0.80 sec after the explosion.Calculate the depth of the ground.
A: 1200 m
B: 2400 m
C: 3200 m
D:5000 m

Answer 1

Answer: The depth of the ground is 3600 m.

Explanation:

Given that,

Distance $d_{1} = 3200\ m$

Detects two signals at $t_{1}$  and $t_{2}$ after the explosion.

$t_{1} = 0.64 sec$

$t_{2} = 0.80 sec$

Now, the speed of sound through the ground

$v= \dfrac{d_{1}}{t_{1}}$

$v = \dfrac{3200\ m}{0.64 s}$

$v = 5000\ m/s$

The second reflection from ground returns after 0.80 sec.

Now, the one side time to the ground is

$t = \dfrac{t_{2}-t_{1}}{2}$

$t = \dfrac{0.80\ s-0.64\ s}{2}$

$t = 0.08 sec$

Now, the depth of the ground is

$d = d_{1} +d_{2}$

$d = d_{1} + vt$

$d = 3200\ m+ 5000\ m/s\times 0.08 s$

$d = 3200\ m+ 400\ m$

$d = 3600\ m$

Hence, the depth of the ground is 3600 m.

Answer 2

Answer:

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Explanation:

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