Question

A man fired a bullet against a wall and heard an echo after $2s$. He walks $80m$ towards the wall and fires a bullet, such that he hears echo after $1s$. Find the distance from the wall to the second fireplace.

Answer 1

Hint:We can approach the solution to this question by simply thinking of it as a problem from the topic distance and displacement instead of reflection of sound. Just consider the velocity of sound as a constant, \[v\] and the distance between the man and the wall in first case as $d$, this makes the distance in second case as $\left( {d\, - \,80} \right)$ and solve the two equations we get to reach to the answer.

Complete step-by-step solution:
We will be trying to approach the solution to the question exactly like we told in the hint section of the solution to the question.
Firstly, we need to discuss the velocity of sound. We know that the medium doesn’t change, so the velocity of sound will be exactly the same in both the cases.
Let’s assume the velocity of sound to be $v\,\,m/s$ and it will stay the same throughout the question.
Now, we will discuss both the cases:
For case A, we have:
Distance between the man and the wall can be assumed to be $d$ meters.
It is given to us in the question that the man hears echo after $2s$, which means that sound takes this much time in travelling twice the distance between the man and the wall.
Now, we can write this situation mathematically as:
$2\, = \,\dfrac{{2d}}{v}$
Now, we will transpose velocity to the other side, by doing this, we will get:
$v\, = \,d$
Now, let’s discuss the second case, case B:
Distance between the man and the wall is now $\left( {d - 80} \right)$ meters since he walked $80m$ towards the wall.
It is given to us that this time, the man hears an echo after $1s$, which means that sound takes this much time in travelling twice the new distance between the man and the wall.
Mathematically, we can represent it as:
$1\, = \,\dfrac{{2\left( {d - 80} \right)}}{v}$
Transposing velocity to the other side, we get:
$v\, = \,2\left( {d - 80} \right)$
We can see that both the cases have an equation for velocity, now, equating both the equations, we get:
$d\, = \,2d - 160$
After transposing, we get:
$d\, = \,160$
But the question has asked us about the new distance, so the answer is actually the value of $\left( {d - 80} \right)$, which is $80m$, which is the correct answer.

Note:- Many students try to also consider the time the bullet takes to reach the wall, which is completely useless in such questions. Other than that, many students solve it correctly but carelessly believe that the answer is the original distance between the man and the wall, while the question has clearly asked about the new distance between them.