Question

A stone is dropped from a height of 20 m. Find the time of reaching at ground and what will be the it velocity before touches the ground​

Answer 1

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$\bf{Given,}$

$\bf \scriptsize{Height=20 m}$

$\bf \scriptsize{Gravitation = 9.8 \: m/ \: sec^{2} }$

$\bf \scriptsize{Initial \: velocity(u) = 0 \: m /\: sec}$

$\bf \scriptsize{Final \: velocity(v) = ?}$

$\bf \scriptsize{Time =?}$

$\bf \scriptsize{Now,}$

$\bf \scriptsize{ \: 1.) \: Using \: relation}$

$\bf \scriptsize{V^{2} = u^{2} + 2gh }$

$\bf \scriptsize{ \implies \: (u)^{2} + 2 \times 9.8 \times 20 }$

$\bf \scriptsize{\implies \: (u)^{2} + 2 \times \frac{98}{ \cancel10} \times \cancel20 }$

$\bf \scriptsize{ \implies \: 0 + 2 \times 98 \times 2}$

$\bf \scriptsize{ \implies \: 4 \times 98}$

$\bf \scriptsize{ \implies \: V^{2} = 392 \: m/sec }$

$\bf \scriptsize{\implies \: v = \sqrt{392} }$

$\bf \scriptsize{ \implies \: V = \sqrt{196 \times 2} } \\ \bf \scriptsize{ = 14 \sqrt{2} \: m \: /sec }$

$\bf \scriptsize{2.) \: For \: a \: freely\: falling \: body.}$

$\bf \scriptsize{V = u + gt}$

$\bf \scriptsize{t \: = \frac{v - u}{g} }$

$\bf \scriptsize{ \implies \: \frac{14 \sqrt{2 - 0} }{9.8} }$

$\bf \scriptsize{ \implies \frac{ \cancel14 \sqrt{2} - 0 \times 10 }{ \cancel98} }$

$\bf \scriptsize{ \implies \: \frac{10 \sqrt{2} }{7} \: sec }$

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$\bf \scriptsize \red{Hence \: final \: velocity \: = 14 \sqrt{2}m / sec \: and \: Time \: is \: \frac{10 \sqrt{2} }{7}sec }$

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