Question

A man fires a gun and hears its echo after 5s. The man then moves 310m towards the hill and fires his gun again. This time he hears the echo after 3s. Calculate the speed of sound.

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Answer 1

Answer:

...

The speed of sound is 310 m/s...

Explanation:

Since the sound will be hearded onlyy when it reflects back so if man moves 310 meter than the sound going and reflecting distance become 2×310=610 and this distance is reduced by 2 second time so speed is 620/2= 310 m/s...

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Speed of Sound has been used. We see that we are given the condition of echo which means the distance covered is twice that is firstly going and returning of echo. Using this we can find the speed of sound. Then we need to find the distance between hill and man. Using this relation we can find the distance between the man and hill.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{v\;=\;\bf{\dfrac{2\:\times\:s}{t}}}}}$

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★ Solution :-

Given,

» Time taken to hear echo for first time = T₁ = 5 s

» Time taken to hear echo for second time = T₂ = 3 s

» Distance moved by man towards hill = 310 m

• Distance between man and hill = s₁ = d

• Distance between man and hill after moving 310 m towards hill = s₂ = (d - 310)

• Let the speed of sound be v which is same for both the cases.

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~ For velocity of sound for First Case ::

For first case we will be using distance as distance between man and hill.

We know that,

$\\\;\sf{:\rightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:s}{t}}}$

• Here s = s₁ and t = t₁

By applying values, we get

$\\\;\sf{:\rightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:s_{1}}{t_{1}}}}$

$\\\;\bf{:\rightarrow\;\;v\;=\;\bf{\green{\dfrac{2\:\times\:d}{5}}}}$

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~ For velocity of sound for the Second Case ::

For second case, the distance will be the new distance between man and hill after subtracting 310 m from it.

We know that,

$\\\;\sf{:\rightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:s}{t}}}$

• Here s = s₂ and t = t₂

By applying values, we get

$\\\;\sf{:\rightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:s_{2}}{t_{2}}}}$

$\\\;\bf{:\rightarrow\;\;v\;=\;\bf{\blue{\dfrac{2\:\times\:(d\:-\:310)}{3}}}}$

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~ For the value of d ::

We know that speed of sound will be same for both cases. So equating both values, we get

$\\\;\bf{\mapsto\;\;\red{v(for\:first\:case)\;=\;\bf{v(for\:second\:case)}}}$

By applying values, we get

$\\\;\bf{\mapsto\;\;\dfrac{2\:\times\:d}{5}\;=\;\dfrac{2\:\times\:(d\:-\:310)}{5}}$

By cross multiplying, we get

$\\\;\bf{\mapsto\;\;3\:\times\:(2\:\times\:d)\;=\;5\:\times\:2\:\times\:(d\:-\:310)}$

$\\\;\bf{\mapsto\;\;6\:\times\:d\;=\;10\:\times\:(d\:-\:310)}$

$\\\;\bf{\mapsto\;\;6d\;=\;10d\;-\;3100}$

$\\\;\bf{\mapsto\;\;-\:10d\;+\;6d\;=\;\;-\;3100}$

$\\\;\bf{\mapsto\;\;-\:4d\;=\;\;-\;3100}$

$\\\;\bf{\mapsto\;\;d\;=\;\;\dfrac{-\;3100}{-\:4}}$

$\\\;\bf{\mapsto\;\;d\;=\;\orange{775\;\;m}}$

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~ For the value of 'v' ::

We already have the equation from first case that,

$\\\;\sf{:\rightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:d}{5}}}$

Now applying the value of d, we get

$\\\;\sf{:\Longrightarrow\;\;v\;=\;\bf{\dfrac{2\:\times\:775}{5}}}$

$\\\;\sf{:\Longrightarrow\;\;v\;=\;\bf{2\:\times\:155}}$

$\\\;\bf{:\Longrightarrow\;\;v\;=\;\bf{\purple{310\;\;m\:sec^{-1}}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:speed\;\:of\;\:sound\;=\;\bf{\purple{310\;\:m\:sec^{-1}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;a\;=\;\dfrac{v\:-\:u}{t}}$

$\\\;\sf{\leadsto\;\;v\:-\:u\;=\;at}$

• This is First Equation of Motion.

$\\\;\sf{\leadsto\;\;s\;=\;ut\;+\;\dfrac{1}{2}\:at^{2}}$

• This is Second Equation of Motion.

$\\\;\sf{\leadsto\;\;v^{2}\;-\;u^{2}\;=\;2as}$

• This is Third Equation of Motion.

$\\\;\sf{\leadsto\;\;v\;=\;\sqrt{2\:gh}}$