A person standing between two vertical cliffs and $480 m$ from the nearest cliff shouts. He hears the first echo after $3 s$ and the second echo $2 s$ later. Calculate the distance of the other cliff from the person.

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Solution

Step 1: Given data
$D_{1}$ is the distance of the person from the nearest cliff and $D_{2}$ is the distance from the farther cliff
$D_{1} = 480 m$
Time after which the first echo is heard, $T_{1} = 3 s$
The time after which a second echo is heard is $T_{2} = 3 + 2 = 5 s$
Step 2: Calculating the speed of sound
We know that the total distance traveled by the sound will be calculated by adding both the directions, in case of going and coming back, for the nearest cliff.
Thus, the total distance that the sound travels before reaching the person is $2 D_{1}$
Let $v$ be the speed of sound
Speed of sound, $v = \frac{2 D_{1}}{T_{1}} = \frac{2 \times 480}{3} = 320 m / s$
Thus, the speed of sound is $320 m / s$ .
Step 3: Calculating the distance of the person from the farther cliff
The sound travels a total distance of $2 D_{2}$ before reaching the person.
$v = \frac{2 D_{2}}{T_{2}}$
$D_{2} = \frac{v \times T_{2}}{2} = \frac{320 \times 5}{2} = 160 \times 5 = 800 m$
Hence, the distance between the other cliff and the person is $800 m$ .

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A person standing between two vertical cliffs and 480m from the nearest cliff shouts. He hears the first echo after 3s and the second echo 2s later. Calculate the distance of the other cliff from the person.

A person standing between two vertical cliffs and $480 m$ from the nearest cliff shouts. He hears the first echo after $3 s$ and the second echo $2 s$ later. Calculate the distance of the other cliff from the person.