A man was standing in front of the verticle cliff $(i)$ he hears the echo of the gun fired by him after $3$ seconds. $(i i)$ he hears the echo of the gun fired by him after $2.5$ seconds when he moves $82.5$ meters closer to the cliff.Find the distance between original position of the man and the cliff.

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Solution

Let the distance between the man and the cliff be $x$ initially.
$∴$ The sound wave travels $2 x$ distance to reach back to the man after $3 s e c$ .
that is, $2 x \to 3 s e c$
or $v = \frac{2 x}{3}$
Now, in $2 n d$ case,
$2 (x - 82.5) \to 2.5$ sec.
or $v = \frac{2 (x - 82.5)}{2.5}$
$∴ \frac{2 x}{3} = \frac{2 (x - 82.5)}{2.5}$
$\frac{5 x}{2} = 3 x - 3 \times (82.5)$
$n = 485 m$

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A man was standing in front of the verticle cliff (i)he hears the echo of the gun fired by him after 3 seconds. (ii)he hears the echo of the gun fired by him after 2.5 seconds when he moves 82.5meters closer to the cliff.Find the distance between original position of the man and the cliff.

Let the distance between the man and the cliff be x initially.∴ The sound wave travels 2x distance to reach back to the man after 3sec.that is, 2x→3sec or v=2x3Now, in 2nd case,2(x−82.5)→2.5sec.or v=2(x−82.5)2.5∴2x3=2(x−82.5)2.55x2=3x−3×(82.5)n=485m