Question

A stone is dropped in a well of depth 19.6m. After how many seconds the sound of splash will be heard to the observer ( Velocity of sound in air = 340 ms ^-1


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Answer 1

$\large\color{#3fff00}\boxed{\colorbox{000080}{Answer : - }}$

Given:

s = 19.6m

V = 340 ms⁻¹

Using second equation of motion:

s = ut + ½ at²

u = 0, a = g

$\color{blue}s = \frac{1}{2} \: {gt}^{2}$

Formula Used :

Second equation of motion s = ut + ½ at²

Time taken to reach the depth:

$t_1 = \sqrt{ \frac{2s}{g} } = \sqrt{ \frac{2 \times 19.6}{9.8} }$

$t_1 = \sqrt \frac{39.2}{9.8}$

$t_1 = \sqrt{4}$

$\color{red}t_1 = 2s$

Time Taken to the sound to reach observer at the top:

$t_2 = \frac{distance}{velocity}$

$= \frac{19.6}{340}$

$= \: 0.05 \: s$

Total time:

$\color{black}\boxed{\colorbox{yellow}{2 + 0.05 = 2.05 s}}$

Answer 2

Answer:

A stone is dropped into a well. The sound of the splash is heard after 2.5s. The depth of the well is 29 m

Explanation:

The sound of splash will be heard after the sound would have travelled the distance equal to the depth of the well.

Let the depth of the well be h

Time taken in reaching the sound to the listener = 2.5 seconds

The velocity of sound in air = 330 m/s

Time taken by the stone in reaching the well

Using the second equation of motion

$\sf{h = ut + \frac{1}{2}gt^{2}}$

$\sf :\longmapsto{h = 0 × t + \frac{1}{2}gt^{2}}$

$\sf :\longmapsto{h = \frac{1}{2}gt^{2}}$

Time taken by the sound of splash to reach the top of the well

$\sf{t' = \frac{h}{330}}$

$\sf{or, \: t' = \frac{gt^{2}}{2 × 330}}$

$\sf{or, \: t' = \frac{10t^{2}}{660}}$ (Taking g = 10 m/s², for simplicity)

$\bf{or, \: t' = \frac{t^{2}}{66}}$

GIVEN:-

$\sf{t + t' = 2.5}$

$\sf :\implies{t + \frac{t^{2}}{66} = 2.5}$

$\sf :\implies{66 + t^{2} = 165 }$

$\sf :\implies{t^{2} + 66t - 165 = 0 }$

$\sf :\implies{t = \dfrac{-66±\sqrt{66^{2} - 4 × (-165) × 1}}{2×1}}$

$\sf :\implies{t = \dfrac{-66±70.82}{2}}$

$\bf \dashrightarrow{t = \cancel{\frac{4.823}{2}}}$ (As time cannot be negative)

$\pink \bigstar\rm\color{blue}\dashrightarrow{t = 2.41 \: second}$

Thus,

$\sf{t' = \frac{2.41^{2}}{66} second}$

$\sf :\longmapsto{t' = 0.09 seconds}$

Therefore, the depth of the well

$\rm{h = 330 × 0.09 \:m}$

$\red \bigstar\bf\color{magenta} \dashrightarrow{h = 29.04m}$

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