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A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after $1.5 s$ and $3.5 s$ respectively, than the distance between the cliffs will be? (Velocity of sound in air $340 m / s$ )

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Detailed Solution

Suppose

d

is the total distance between two cliffs and

x

is the distance between the man and one cliff.

2 x = v \times t_{1}

\Rightarrow x = \frac{v \times t_{1}}{2}

\Rightarrow x = \frac{340 \times 1.5}{2}

\Rightarrow x = 255 m

distance between the man and Second cliff.

2 (d - x) = v \times t_{2}

\Rightarrow (d - x) = \frac{v \times t_{2}}{2}

\Rightarrow (d - x) = \frac{v \times t_{2}}{2}

\Rightarrow (d - x) = \frac{340 \times 3.5}{2}

\Rightarrow (d - x) = 595 m

So, Total distance

d = 595 + x

= 595 + 255

= 850 m

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