Question

a person standing between tow vertical clifffs and 640m form the nearest cliff producses sound.He hers the first echo after 4s and the second echo 3s later.calculate 1. speed of sound in air 2.the distance between the cliffs

Answer 1

Question :

A person standing between two vertical cliffs and 640m form the nearest cliff produces sound.He hears the first echo after 4s and the second echo 3s later.calculate

1. Speed of sound in air

2. The distance between the cliffs.

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GiveN :

• A person is standing between two cliffs.
• Distance from nearest cliff (D1) = 640 m
• echo heard from nearest cliff (t1) = 4 seconds

And for Far Cliff (2nd cliff)

• Distance = ?
• Time (t2) = 4 + 3 = 7 seconds

Time is added because 2nd echo is heard after 3 seconds from 1st one.

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To FinD :

• Speed of sound
• Distance between cliffs

SolutioN :

Use formula for Speed for the nearest cliff from the man.

⇒V = Distance/Time

⇒V = 2(D1)/Time

✯Here, distance is taken double because first person speaks and then echo will be heard from cliff. So, sound have to travel double distance.

⇒V = 2(640)/4

⇒V = 1280/4

⇒V = 320

∴ Speed of sound is 320 m/s

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Now, distance between the cliffs.

⇒V = 2(D2)/time

Velocity will be same as the nearest cliff, and time will be 7 seconds.

⇒320 = 2(D2)/7

⇒2(D2) = 320 * 7

⇒2(D2) = 2240

⇒D2 = 2240/2

⇒D2 = 1120

∴ Distance from 2nd Cliff (far cliff) from man is 1120 m

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Now, for total distance add both the distance of cliffs from that of man

⇒D = D1 + D2

⇒D = 640 + 1120

⇒D = 1760

∴ Distance between cliffs is 1760 m

Answer 2

$\large{\pink{\bold{\underline{\underline {Figure:}}}}}$

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$\large{\red{\bold{\underline{\underline{Answer}}}}}$

$\purple{\sf{\mapsto Speed \: of \:sound\: in \:air = 320 \: m/s}}\\ \\\purple{\sf{\mapsto Distance between the cliffs = 1760 \: m}}$

$\bold{\green{\underline{Given}}} \\ \\ \sf{\rightsquigarrow Distance\: (d_1) = 640 \: m} \\ \\ \sf{\rightsquigarrow Time \: taken \: (t_1) =4 \: s } \\ \\ \bold{\blue{\underline{To \: Find}}} \\ \\ \sf{\rightsquigarrow Speed \: of \: sound \:in \: air =\:?}\\ \\ \sf{\rightsquigarrow Distance \: between \: the \: cliffs =\:?}$

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❏ According To Given Question

⠀⠀

$\dashrightarrow \sf{t_2= 4+3=7\:s}\\ \\ \sf{First\: echo \:heard \:from\: the\: nearest \:cliffs.}\\ \\ \sf{Total\: distance\: travelled = 2 d_1 = 2\times 640} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1280\: m/s} \\ \\ \sf{: \implies v = \frac{2 \: d_1}{t_1} } \\ \\ \sf{ : \implies v = \frac{ \cancel{1280}}{ \cancel{4} }} \\ \\ \sf{ \purple{ : \implies}}\purple{\underline{\boxed{\sf{ v = 320 \: m/s}}}}\\ \\ \sf{Second \:echo \:is \:heard\: to \:secod\: cliffs} \\ \\ \sf{ : \implies v = \frac{2 \: d_2}{t_2} } \\ \\ \sf{ : \implies 320 = \frac{2 \: d_2}{t_2} } \\ \\ \sf{ : \implies d_2 = \frac{320 \times 7}{2} } \\ \\ \sf{ : \implies d_2 = \frac{ \cancel{2240}}{ \cancel{2} }} \\ \\ \sf{ : \implies d_2 = 1120 \: m} \\ \\ \sf{Hence,\: the\: distance\: between\: two\: cliffs=d_1+d_2}\\ \\ \sf{ : \implies 640 + 1120} \\ \\ \sf \purple{: \implies \underline{ \boxed{ \sf{1760 \: m}}}}$