Question

A boy dropped a stone in a river from the height of 40 m . if the speed of sound is 344m/s . then after how much time will he hear the splash ?​

Answer 1

Answer :-

The boy will hear the splash after 2.96 seconds .

Explanation :-

We have :-

→ Height = 40 m

→ Initial velocity of stone (u) = 0 m/s

→ Speed of sound in air = 344 m/s

→ Gravitational acceleration (g) = 9.8 m/s²

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Firstly, let's calculate the time taken by the stone to reach the surface of the river by using the 2nd equation of motion.

h = ut + ½gt²

⇒ 40 = 0(t) + ½ × 9.8 × t²

⇒ 40 = 4.9t²

⇒ t² = 40/4.9

⇒ t = √8.16

⇒ t = 2.85 s

Now, we have to calculate the time taken by the splash to reach the boy.

Time = Distance/Speed

⇒ t' = 40/344

⇒ t' = 0.11 s

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Thus, total time taken by the boy to hear the splash is :-

= t + t'

= (2.85 + 0.11) s

= 2.96 s

Answer 2

Answer:

Given :-

• A boy dropped a stone in a river from the height of 40 m.
• The speed of sound is 344 m/s.

To Find :-

• How much time will he hear the splash.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} gt^2}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time
• g = Acceleration due to gravity

$\clubsuit$ Speed Formula :

$\mapsto \sf\boxed{\bold{\pink{Time =\: \dfrac{Distance}{Time}}}}$

Solution :-

First, we have to find the time taken by the stone to reached the surface of water :

${\small{\bold{\purple{\underline{\bigstar\: In\: first\: case\: :-}}}}}$

$\mapato$ A boy dropped a stone in a river from the height of 40 m.

Given :

• Initial Velocity = 0 m/s
• Distance Covered = 40 m
• Acceleration due to gravity = 9.8 m/s²

According to the question by using the second equation of motion formula we get,

$\implies \sf 40 =\: (0)t + \dfrac{1}{2} \times 9.8 \times t^2$

$\implies \sf 40 =\: 0 + \dfrac{1}{2} \times \dfrac{98}{10} \times t^2$

$\implies \sf 40 =\: 0 + \dfrac{98}{20} \times t^2$

$\implies \sf 40 =\: \dfrac{98}{20} \times t^2$

$\implies \sf 40 \times \dfrac{20}{98} =\: t^2$

$\implies \sf \dfrac{800}{98} =\: t^2$

$\implies \sf 8.16 =\: t^2$

$\implies \sf \sqrt{8.16} =\: t$

$\implies \sf 2.85 =\: t$

$\implies \sf\bold{\green{t =\: 2.85\: seconds}}$

Therefore, the time taken by the stone to reached the surface of water is 2.85 seconds .

Again, we have to find the time taken by the sound to reached the top of water :

${\small{\bold{\purple{\underline{\bigstar\: In\: second\: case\: :-}}}}}$

$\mapsto$ The speed of sound is 344 m/s.

Let,

$\leadsto$ The time be t seconds

Given :

• Distance Covered = 40 m
• Speed = 344 m/s

According to the question by using the speed formula we get,

$\implies \sf 344 =\: \dfrac{40}{t}$

By doing cross multiplication we get,

$\implies \sf 344t =\: 40$

$\implies \sf t =\: \dfrac{\cancel{40}}{\cancel{344}}$

$\implies \sf\bold{\green{t =\: 0.16\: seconds}}$

Hence, the time taken by the sound to reached the top of water is 0.11 seconds.

Now, we have to find the time will he hear the splash :

$\longrightarrow \sf Time\: after\: which\: splash\: is\: heard =\: 2.85 + 0.11\: seconds$

$\longrightarrow \sf Time\: after\: which\: splash\: is\: heard =\: \dfrac{285}{100} + \dfrac{11}{100}\: seconds$

$\longrightarrow \sf Time\: after\: which\: splash\: is\: heard =\: \dfrac{285 + 11}{100}\: seconds$

$\longrightarrow \sf\bold{\red{ Time\: after\: which\: splash\: is\: heard =\: 2.96\: seconds}}$

$\therefore$ The time will he hear the splash is 2.96 seconds.