A stone is dropped into a well of depth 125 metres. Calculate the time

1.	A stone is dropped into a well of depth 125 metres. Calculate the time taken by the sound of the splash to reach the man.
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{Time\:taken=5.4\:sec}}}$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies Depth \: of \: stone = 125 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Time \: taken \: by \: sound \: of \: splash \: to \: reach \: the \: men =?$ • ACCORDING to given QUESTION : $\tt \circ \: Initial \: speed = 0 \: m/s \\ \\ \tt \circ \: Acceleration = 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} at^{2} \\ \\ \tt: \implies 125 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \tt: \implies 125 \times 2 = 10 \times {t }^{2} \\ \\ \tt: \implies \frac{250}{10} = {t}^{2} \\ \\ \tt: \implies {t}^{2} = 25 \\ \\ \tt: \implies t = \sqrt{25} \\ \\ \green{\tt: \implies t_{1} = 5 \: sec} \\\\ \bold{As \: we \: know \: that} \\ \tt: \implies Time \: taken \: by \: sound = \frac{125}{340} \\ \\ \tt: \implies t_{2} = 0.367 \: sec\\\\ \tt:\implies Total\:time=t_{1}+t_{2}\\\\\tt:\implies Total\:time=5+0.4\\\\\tt:\implies Total\:time=5.4\:sec \\\\ \green{\tt \therefore After \: 5.4 \: sec \: of \: stone \: dropped \: the \: sound \: of \: splash \: reach \: the\: man} \\ \\ \blue{ \bold{Some \: related \: formula}}\\\orange{\tt\circ\:v=u+at}\\\\ \orange{\tt\circ\:v^{2}=u^{2}+2as}$

A stone is dropped into a well of depth 125 metres. Calculate the time taken by the sound of the splash to reach the man.

Answer»

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Time\:taken=5.4\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies Depth \: of \: stone = 125 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Time \: taken \: by \: sound \: of \: splash \: to \: reach \: the \: men =?$

• ACCORDING to given QUESTION :

$\tt \circ \: Initial \: speed = 0 \: m/s \\ \\ \tt \circ \: Acceleration = 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} at^{2} \\ \\ \tt: \implies 125 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \tt: \implies 125 \times 2 = 10 \times {t }^{2} \\ \\ \tt: \implies \frac{250}{10} = {t}^{2} \\ \\ \tt: \implies {t}^{2} = 25 \\ \\ \tt: \implies t = \sqrt{25} \\ \\ \green{\tt: \implies t_{1} = 5 \: sec} \\\\ \bold{As \: we \: know \: that} \\ \tt: \implies Time \: taken \: by \: sound = \frac{125}{340} \\ \\ \tt: \implies t_{2} = 0.367 \: sec\\\\ \tt:\implies Total\:time=t_{1}+t_{2}\\\\\tt:\implies Total\:time=5+0.4\\\\\tt:\implies Total\:time=5.4\:sec \\\\ \green{\tt \therefore After \: 5.4 \: sec \: of \: stone \: dropped \: the \: sound \: of \: splash \: reach \: the\: man} \\ \\ \blue{ \bold{Some \: related \: formula}}\\\orange{\tt\circ\:v=u+at}\\\\ \orange{\tt\circ\:v^{2}=u^{2}+2as}$

A stone is dropped into a well of depth 125 metres. Calculate the time taken by the sound of the splash to reach the man.

• ACCORDING to given QUESTION :

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment