Question

A man fired a bullet in front of a mountain and he heard its echo after 2 seconds. After travelling a distance 85 m towards the mountain, fired another bullet and heard its echo after 1.5 seconds. The velocity of the sound and distance between the mountain and the man when the first bullet was fired are
(A) \[340\,m/s, 340\,m\]
(B) \[340\,m/s, 140\,m\]
(C) \[140\,m/s, 340\,m\]
(D) \[140\,m/s, 140\,m\]

Answer 1

Hint: Using the relation between velocity, distance and time express the distance between the mountain and the person after the first shot and after the second shot. Solve the two equations you obtained to get the distance between the mountain and the person.

Formula used:
\[v = \dfrac{d}{t}\]
Here, v is the velocity, d is the distance and t is the time.

Complete step by step answer:
We assume the distance between the mountain and the man when he fired the first shot is d. We know that for an echo, the distance travelled by the sound wave is twice the distance between the source and point of reflection of the wave.
We can express the distance between the mountain and the man for the first echo as follows,
\[2d = v{t_1}\]
Here, v is the velocity of sound and \[{t_1}\] is the time for the first echo.
Substituting 2 second for \[{t_1}\] in the above equation, we get,
\[2d = 2v\]
\[ \Rightarrow d = v\] …… (1)
Now, for the second shot, we have given that the man travels 85 m distance in the direction of the mountain. Therefore, for the second shot, the distance between the mountain and the man is\[\left( {d - 85} \right){\text{m}}\].
We can express the distance between the mountain and the man for the second echo as follows,
\[2\left( {d - 85} \right) = v{t_2}\]
Here, \[{t_2}\] is the time for the second echo.
Substituting 1.5 second for \[{t_2}\] in the above equation, we get,
\[2\left( {d - 85} \right) = 1.5v\]
From equation (1), we have, \[d = v\]. Therefore, the above equation becomes,
\[2\left( {d - 85} \right) = 1.5d\]
\[ \Rightarrow 2d - 170 = 1.5d\]
\[ \Rightarrow 2d - 1.5d = 170\]
\[ \Rightarrow \dfrac{1}{2}d = 170\]
\[ \therefore d = 340\,m\]
Therefore, the distance between the mountain and the man is \[340\,m\] and the velocity of the sound from equation (1) is also \[340\,m/s\].

So, the correct answer is option (A).

Note: To answer this question theoretically, we could have used the fact that the velocity of sound in air is constant and it is 340 m/s. Therefore, you can use the relation between distance, velocity and time to determine the distance between the mountain and the man after the first shot. Remember, the distance travelled by the sound for an echo is always twice the distance between the source and point of reflection of the sound.