Question

A man stands in front of a cliff and fires a gun. He hears an echo after 4 seconds. on moving closer to the cliff by 41.25 m and again firing a gun, he hears an echo after 3.75 s.​. find the distance of man from final position

Answer 1

Answer:

The final position of the man from cliff is 618.75m

Explanation:

Let the distance between the initial position of the man and the cliff is $x$.

Time taken to hear the first echo = 4s

The distance between the final position of the man and the cliff = $x - 41.25 m$.

Time taken to hear the second echo = 3.75s

We know that ,

                        Speed = $\frac{distance}{time}$

                           $v=\frac{2x}{t}$   {∵ the total distance travelled by the sound is 2x}

                            $v = \frac{2x}{4}$ →(1)

 When the man moved a distance of 41.25m  closer to the cliff, the equation for speed becomes,

                            $v = \frac{2(x-41.25)}{3.75}$ → (2)

From equations (1) & (2),

                             $\frac{2x}{4} = \frac{2(x-41.25)}{3.75}$

                            $\frac{2x}{4}\times 3.75 = 2x - (2\times 41.25)$

                            ⇒ $x = 660 m$

∴ The final position of the man from cliff = $x - 41.25 = 660 - 41.25 = 618.75 m$