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** **__Solve this__ ** **__:__

A person standing between two vertical cliffs and 640 m away from the nearest cliff shouted. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate.

(a) the velocity of sound in air, and

(b) the distance between the cliffs

__Solve this__

__:__

$\left(i\right)Velocityofthesoundv=\frac{2d}{t}\phantom{\rule{0ex}{0ex}}v=\frac{2\times 640}{4}=320m/s\phantom{\rule{0ex}{0ex}}\left(ii\right)Dis\mathrm{tan}cebetweenthetwocliffs\phantom{\rule{0ex}{0ex}}v=\frac{2x\text{'}}{t},totaltimeis7sec\phantom{\rule{0ex}{0ex}}or320=\frac{2x\text{'}}{7}\phantom{\rule{0ex}{0ex}}x\text{'}=1120m\phantom{\rule{0ex}{0ex}}Dis\mathrm{tan}cebetweenthetwocliffs\phantom{\rule{0ex}{0ex}}sox=1120+640=1760m$

Regards

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