Question

If a splash is heard 4.23second after a stone is dropped into the well 78.4m find the speed of air

Answer 1

Given:

Depth of well (h) = 78.4 m

Time at which splash is heard after dropping stone (t) = 4.23 s

To Find:

Speed of sound in air (v)

Answer:

Let time taken by stone be $\sf t_1$ seconds to strike the water surface of well after being released from a height (h)

$\rm \implies h = \dfrac{1}{2}g t_1 ^{2} \\ \\ \rm \implies t_1 = \sqrt{ \dfrac{2h}{g} }$

Let the sound of splash be heard $\sf t_2$ seconds after the stone strikes the water surface. Then,

$\rm \implies t_2 = \dfrac{h}{v}$

Total time i.e. Time at which splash is heard after dropping stone;

$\rm t = t_1 + t_2 = 4.23 \ s$

So,

$\rm \leadsto \sqrt{ \dfrac{2h}{g} } + \dfrac{h}{v} = 4.23 \\ \\ \rm \leadsto \sqrt{ \dfrac{2 \times 78.4}{9.8} } + \dfrac{78.4}{v} = 4.23 \\ \\ \rm \leadsto \sqrt{2 \times 8} + \dfrac{78.4}{v} = 4.23 \\ \\ \rm \leadsto \sqrt{16} + \dfrac{78.4}{v} = 4.23 \\ \\ \rm \leadsto 4 + \dfrac{78.4}{v} = 4.23 \\ \\ \rm \leadsto \dfrac{78.4}{v} = 4.23 - 4 \\ \\ \rm \leadsto \dfrac{78.4}{v} = 0.23 \\ \\ \rm \leadsto v = \dfrac{78.4}{0.23} \\ \\ \rm \leadsto v = 340.87 \: m {s}^{ - 1}$

$\therefore$ Speed of sound in air (v) = 340.87 m/s

Answer 2

Answer :

›»› The speed of air is 340.87 m/s.

Given :

• If a splash is heard 4.23 second after a stone is dropped into the well 78.4 m.

To Find :

• The speed of air.

Solution :

Let us assume that, the time taken by stone to hit the water surface is the well is t₁ seconds.

And, let us assume that, the time taken be splash of sound to reach the top of well is t₂ seconds.

Total time taken,

→ t = t₁ + t₂

→ t = 4.23 sec

Now, for downward journey of stone,

• Initial velocity (u) = 0 m/s.
• Acceleration due to gravity (g) = 9.8 m/s².
• Stone is dropped into the well (h) = 78.4 m.
• t = t₁ = ?

From second equation of motion

$\tt{:\implies h = ut + \dfrac{1}{2}g{t}^{2} }$

⠀

$\tt{:\implies 78.4 = 0 + \dfrac{1}{2} \times 9.8 \times {t}_{1}^{2} }$

⠀

$\tt{:\implies 78.4 = 0 + 1 \times 4.9 \times t_{1}^{2} }$

⠀

$\tt{:\implies 78.4 = 0 + 4.9 \times t_{1}^{2} }$

⠀

$\tt{:\implies 78.4 = 0 + 4.9t_{1}^{2} }$

⠀

$\tt{:\implies 78.4 = 4.9t_{1}^{2} }$

⠀

$\tt{:\implies t_{1}^{2} = \dfrac{78.4}{4.9} }$

⠀

$\tt{:\implies t_{1}^{2} = 16}$

⠀

$\tt{:\implies t_{1} = \sqrt{16} }$

⠀

$\bf{:\implies t_{1} = 4}$

⠀

Now,

→ t₁ + t₂ = 4.23

→ 4 + t₂ = 4.23

→ t₂ = 4.23 - 4

→ t₂ = 0.23 sec

⠀

If velocity of sound in air,

$\tt{:\implies Velocity = \dfrac{Height}{Time \: taken}}$

⠀

$\tt{:\implies Velocity = \dfrac{78.4}{0.23}}$

⠀

$:\implies \red{\boxed{ \blue{\textsf{\textbf{Velocity = 340.87 m/s}} }}}$

⠀

Hence, the speed of air is 340.87 m/s.