Question

A man standing in front of a vertical cliff fires a gun. he hears the echo after 3.5 seconds. on moving a closer to the cliff by 84m , he hears the echi after 3 seconds. calculate the distance of the cliff from the initial position of the man.​

Answer 1

Answer:

$588 metres.$

Explanation:

$d_{1} = x$                                                            $d_{2} = x - 84$

$t_{1} = 3.5 s$                                                        $t_{2} = 3s$

                              $V_{1} =V_{2}$

$\frac{2d_{1}}{t_{1} } = \frac{2d_{2}}{t_{2} }$

$\frac{2x}{3.5} = \frac{2(x-84)}{3}$

$6x = 3.5*2(x-84)$

$6x=7(x-84)$

$6x=7x- 588$

$7x-6x=588$

$x= 588 m$

This was the question of today's physics paper. Even I am from 10th I.C.S.E. Hope it was nice for you. All the best for the next Chemistry exam:) and also if this answer was helpful, please mark it as the brainliest:)

Answer 2

Answer:

588 m

Explanation:

Let the distance of the cliff from initial position be d m and speed of sound be V m/s

For first echo, t = 2d/V = 3.5 s -(1)

let distance of cliff from initial position be (d - 84)

Therefore, for second echo, t= 2(d-84)/V = 3s -(2)

Dividing (1) and (2)

d/(d-84) = 3.5/3

cross mutiply, we get 3d = 3.5d - 294

294 = 0.5d

d= 588m