Question

A man fired a bullet against a wall and hears an echo after 2 s. He walks 80m towards the wall and fired the bullet, such that he hears echo after 1s.The distance from wall to his first position is

Answer 1

The distance from wall to his first position is 160m

Answer 2

$\sf\huge\bold{Answer:}$

From the given data we have,

$d _{walk}$=80m

$t _{1}$=2s

$t_{2}$=1s

$d _{2}$=?

v= velocity of sound

When he fires the bullet, the sound generated travels everywhere, including in the direction towards the wall. 

The sound wave travelling towards the wall, hits it, gets reflected from it, moves backwards and reaches his ear.

This is how he hears the echo.

This shows that the sound wave travels twice the distance between him and the wall. Hence, while calculating, we'll have to double the distances.

$⟶2d _{1} = v \times t _{1}...(i)$

$⟶2d _{2} = v \times t _{2}...(ii)$

$⟶d _{1} = d _{2} + d _{walk}...(iii)$

Step 1 : Diving equation ii by i

$⟶ \frac{d _{2}}{d _{1}} = \frac{t _{2}}{t _{1}} ...(iv)$

Step 2 : Substituting 3 in 4

$⟶\frac{d _{2}}{d _{2} +d _{walk} } = \frac{t _{2}}{t _{1}}$

$⟶ \frac{d _{2}}{d _{2} +80 } = \frac{1}{2}$

$∴2d _{2} = d _{2} + 80$

$∴d _{2} = 80m$

⇒ the distance from wall to the second fired place is 80 metres.