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Question
A person standing between two vertical cliffs and 480 m from the nearest cliff shouts. He hears the first echo after 3s and the second echo 2s later. Calculate:
- The speed of sound.
- The distance of the other cliff from the person
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Solution
Given: Distance of nearest cliff, \(d_1 = 480\ \text{m}\)
Time for first echo, \(t_1 = 3\ \text{s}\)
(i) For an echo, sound travels to the cliff and back:
\[ v=\frac{2d_1}{t_1} \]
\[ v=\frac{2\times480}{3}\]
\[ {v=320\ \text{m/s}} \]
(ii) The second echo is heard 2 s after the first echo.
\[ t_2=3+2=5\ \text{s} \]
Let the distance of the other cliff be \(d_2\).
\[ d_2=\frac{vt_2}{2} \]
\[ d_2=\frac{320\times5}{2}\]
\[{d_2=800\ \text{m}} \]
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