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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 distance moved by the person.

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ICSE Sp 2024

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Answer

(a) Let the distance of the cliff from the initial position of the person be x m.

Time after which echo is heard = 3 sec

So, the distance travelled by sound in 3 sec = 2x m

Speed of sound in air = 336 m/s

Distance = speed x time

Substituting the values we get,

2x=336×3x=336×32x=168×3x=504 m2x = 336 \times 3 \\[1em] \Rightarrow x = \dfrac{336 \times 3}{2} \\[1em] \Rightarrow x = 168 \times 3 \\[1em] \Rightarrow x = 504 \text{ m}

∴ The initial distance of the person from the cliff is 504 m.

(b) As time taken to hear the echo is increased by 1.5 sec, hence, the person has moved away from the cliff.

Let the distance moved by the person from his initial position be y m.

Time after which echo is heard = 1.5 + 3 = 4.5 sec

New distance = 504 + y

So, the distance travelled by sound in 4.5 sec = 2(504 + y) m

Speed of sound in air = 336 m/s

Distance = speed x time

Substituting the values we get,

2(504+y)=336×4.5504+y=336×4.52y=(168×4.5)504y=756504y=252 m2(504 + y) = 336 \times 4.5 \\[1em] \Rightarrow 504 + y = \dfrac{336 \times 4.5}{2} \\[1em] \Rightarrow y = (168 \times 4.5) - 504 \\[1em] \Rightarrow y = 756 - 504 \\[1em] \Rightarrow y = 252 \text{ m}

∴ Distance moved by the person = 252 m

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