A stone is dropped into a well of depth 20 m. If the splash of water is heard after 2.06 sec. Find the

velocity of sou

2 thoughts on “A stone is dropped into a well of depth 20 m. If the splash of water is heard after 2.06 sec. Find the <br /><br />velocity of sou”

1. Answer:

   what you are telling do you have sense or not

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2. [tex] \frak{given}\begin{cases} \cdot \sf \: total \: time = \frak{2.06 \: sec} \\ \cdot \: \sf depth = \frak{20 \: m} \: \\ \cdot \sf \: initial \: velocity(u) = \frak{0 \: m {s}^{ – 1} } \end{cases}[/tex]

   First we are gonna calculate the time taken by stone.

   Then subtract that time by the total time given in order to find the time taken by sound.

   time taken by the stone can be calculated by the following formula:-

   [tex] \rightarrow \large \underline {\boxed{ \pink{{ \frak{s = ut + \dfrac{1}{2}g {t}^{2} }}}}} \bigstar[/tex]

   [tex] \: \: \: \: \: \: \: \: \dag\underline \frak{substituting \: the \: given \: values \: in \: the \: formula}[/tex]

   [tex] : \implies \sf 20 = (0) t + \dfrac{1}{2} \times 10 \times {t}^{2} [/tex]

   [tex] : \implies \sf 20 = 0 + 5 {t}^{2} [/tex]

   [tex] : \implies \sf {t}^{2} = \dfrac{ \cancel{20}}{ \cancel{5} } = 4[/tex]

   [tex] \ : \implies { \boxed { \pink{{\frak{t = 2 \: sec}}}}} \bigstar[/tex]

   [tex] \sf time \: taken_{(stone)} = total \: time – time \: taken_{(sound)} [/tex]

   [tex] : \implies \sf time \: taken _{(stone)} = 2.06 – 2[/tex]

   [tex] : \implies \sf time \: taken_{(stone)} = 0.06 \: sec[/tex]

   Finally, we are gonna calculate the value of velocity of sound in air by using the formula:-

   [tex] \rightarrow \large\underline { \boxed {\pink{{ \frak{velocity = \frac{displacement}{time} }}}}} \bigstar[/tex]

   [tex] : \implies \sf velocity_{(sound)} = \dfrac{20}{0.06} [/tex]

   [tex] : \implies \boxed { \pink{{\frak{ velocity_{(sound)} = 333.3 \: m {s}^{ – 1} }}}} \bigstar[/tex]

   Hope it’s beneficial 😀

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