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Physics

A person standing in front of a cliff fires a gun and hears its echo after 3s. If the speed of sound in air is 336 ms-1.

(a) Calculate the distance of the person from the cliff.

(b) After moving a certain distance from the cliff, he fires the gun again and this time the echo is heard 1.5 s later than the first. Calculate the distance that the person moved.

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Answer

(a) Given,

Time taken to hear the echo = 3s

Speed of sound = 336 ms-1

So,

Distance of the person from the cliff=Speed of sound×Time taken2=336×32=10082=504 m\text {Distance of the person from the cliff} = \dfrac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1em] =\dfrac{336 \times 3}{2} = \dfrac{1008}{2} \\[1em] = 504\ \text m

So, distance of the person from the cliff is 504 m.

(b) Given,

Time taken to hear the new echo = 3 s + 1.5 s = 4.5 s

So,

New distance of the person from the cliff=Speed of sound×Time taken2=336×4.52=15122=756 m\text {New distance of the person from the cliff} = \dfrac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1em] =\dfrac{336 \times 4.5}{2} = \dfrac{1512}{2} \\[1em] = 756\ \text m

Distance moved by the person = 756 - 504 = 252 m

So, distance moved by the person is 252 m.

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